6 to the (3x + 5) = 1, many don’t know where to start
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- Опубликовано: 1 окт 2024
- How to solve an exponential equation using logarithms.
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Start with the 1. 1 = 6⁰, so 6³ˣ⁺⁵ = 6⁰ equates to 3x + 5 = 0, and x = -5/3.
Yes, I saw this in a couple of seconds and it looks quite ridiculous to use logarithms here, except if the point was just training logarithms.
Well, Einstein said it well:
"Always describe a problem as simply as possible, but no simpler."
Same for programming, don't reinvent the wheel--unless you discover a better wheel."
@@BluesChoker01do you have any idea on where to start with programming…I tried to instal the python app in my laptop but it declined now I don’t know what to do from here ..
Yes the simplest and logically correct।
That's also how I solved it. But I guess he wants to teach the general method.
Don't need to use logarithms... formally, at least. You just need to remember the zero exponent rule.
Just make the bases the same. By the zero exponent rule, 1 can be rewritten as 6^0.
6^(3x+5) = 6^0 => since the bases are equal, then the exponents must be equal.
3x+5 = 0 => x=-5/3
I said x=-(5/3) within 3 seconds. It’s very straightforward when you know n^0=1 for finite n≠0.
Yes, I think this a very roundabout way of doing this. He should remember that math tests have time limits.
@@paullambert8701 Definitely the quickest way when we know n^0 = 1, but if it is not =1 how do you solve?
@@johnwythe1409 :Yes, this only works if YadaYada =1. If not, then :(((
Well aren't you just the smartest boy?
@@paullambert8701 If you use logarithms it is also fairly quick. Reason it took so long in the video is because he spent several minutes teaching what logarithms are and how to use them.
Fast take - this is a thing-to-zero-power because the result is 1. That means whatever it takes to get 6 to the 0 power. Thus 0=3x+5... ergo -5=3x so x=-3/5
Thank you!
How I did it too.
Your fast take was wrong. You put x=-3/5; the correct answer was x=-5/3. So the moral to the story is to slow down.
Precisely,,
ERROR FINAL!
Greetings. The answer is negative 5/3. The expression 6^(3X+5)=1 can
be rewritten as 6^(3X+5)=6^0.
Therefore, 3X+5=0 by equating the values of the exponents. Now, finding the value for X is simple. We will transpose the known value 5 across the sign of equality to get
3X=-5, and X=-5/3 after dividing both sides by 3. Lovely.
Even less work than my solution - good job!
@@andrewhines1054 Greetings. Blessings.
That’s how I did it.
Good, correct, and detailed answer, Devon. I am a down, dirty, and done type of guy: (in my head) "Anything raised to the 0 power equals 1, so 3x+5 = 0, x=-5/3, next?)
If Math Man John wanted to illustrate how to manipulate exponents and logarithms to solve the more generalized problems as he does in his explanation, he should have used a different problem. By using "1" here, he invites the analysis you and I did, but if it had read "... = 28", the "exponent is 0" trick wouldn't be useful.
👍
I'm really sorry, but you need a lot of words to explain stuff. I lost all interest before you came even close to solving the problem.
6^(3x+5)= 6^0
3x+5=0
x=-5/3
I'm 72, previously an engineering student, and sea captain. Did the sea captain thing after enginering calculus without a calculator got the best of me. Only to be confronted with the joys of spherical trigonometry that is required to do celestial navigation. I'm just saying you younguns with your GPSs and calculators don't know how easy you've got it. I watch your program to help me from forgeting everything I ever learned.
But the Earth isn't a perfect sphere. There'll be errors in navigation that needs adjustments to follow the appropriate path to the next port. Otherwise, you could encounter something unexpected and bad.
@@oahuhawaii2141 Which is what they do. Since it's very close the spherical trig gets you quite close to where you need to be and then you can make the adjustments that is part of the navigation process.
If you need to navigate at sea, you still need to be able to find your location the old way. Your GPS may be broken or fell overboard.
Its a easy solution
Take 1 = 6^0 bcoz anything power to 0 is 1
Therefore
3x+5 =0
X= -5/3
No need for this long process
Why make it so difficult? The power must be 0 for the answer to be 1. So 3x+5=0; x=-5/3
They always drill exponents long time before they even whisper of a log
Any number to 0 power is 1 3x+5=0 done.
So it would seem that many of you are brilliant students who were able to recognize that any number to the zero power is equal to 1 and were able to solve this problem in record time... bravo. What you missed was the brilliance of the teacher to use this problem to be solved using the rules of logarithms, that many may not know or possibly are rusty with, and through his steps have given the student an opportunity to see that these crazy rules actually work and the result actually comes out to a value that we can accept and know to be true. The actual student now has validation of the methods that have been shown result in an answer that they can validate.
I like the detail you went into here. I think people who want, or need, context will appreciate it, too. Everyone else probably just wants a quick solution or trick. No dig at them, they probably know most of it. I'm just saying not to be discouraged by people who want you to rush. Videos like this help many who need it.
In this particular case no need to unneccesry a long and q boring method।time is most important here also।
@@girdharilalverma6452 Your answer is incomplete and grammatically incorrect.
I learned logs back in the day but I didn't even go that long route to the problem, because I also learned that any number to the zeroth power equals 1, so I solved this in my head by just thinking "3x-5=0", and then solving for x - simple math, without using a calculator or log table. I don't care what the base number is.
This is a funny one but I saw the trick: any number to the power zero (0) = 1, and here we have such a power: 3X + 5 = 0 => 3X = - 5 => X = - 5/3. That way we get 6^0 = 1. :)
The log of a number is the *_power_* to which a *_base_* must be raised to give the number. You can iinvent your own base, but using the example given, log (base 2) of 16 = 4, because 4 is the power to which the base (2) must be raised to give 16.
Here we write that as follows: 2log16 = 4 (because, as you already said, 2^4 = 16; (bacon = base: 2), eggs = result: 16), and / answer: 4).
I typed into my calculator 'log10÷log4 and the answer was 1.22 weekday am i doing wrong?
Using log to base 10 ( the normal log button):
log 10 = 1
log 4 = 0.6 approx
gives 1.66 when divided
Using natural log (the ln button):
log 10 = 2.3 approx
log 4 = 1.386 approx
again gives 1.66
I've no idea how you arrived at 1.22 (even tried mixing the two log functions).
edit - I used your 1.22 and swapped around to find either:
log 10/log 6.6 = 1.22
or
log 5.426/log 4 = 1.22
Doesn't help much!
On the Windows calculator (scientific mode)
Simply input 10 press log - then divide by 4 press log
i.e. (10 log)/(4 log)
Not sure what calculator you are using.
…may depend on the hierarchical structure (program) built into your calculator. On mine I hit 10 (the 1 and 0 buttons), then hit the log 10 button, then hit the divide button, then hit the 4 button, then hit the log10 button, then hit the = button and the calculator gives the approximation 1.66……….. how you enter those terms in the proper sequence and correctly on a calculator depends on the calculator’s hierarchy structure.
No need for log. Any number to the zero power is always 1. So, 3x+5=0. 3x is -5. Divide bot side by 3. X equals -5
Yes many won’t no where to start because they preoccupied by the equation of how to make there paycheck equal to the cost of living.
Step #1, convert 1 into 6^0, then with the same base, exponents must be equal.
@Mike-lx9qn Yes, but it seems unnecessarily complex and does not teach the principle that when bases are equal, then the exponents must be equal.
@@jamesharmon4994 "bases are equal, then the exponents must be equal."
However, 5^0 = 1, 6^0 also = 1, so it is not clear that one must choose 6^0. Instead it is probably sufficient to know that any number to the 0 power is 1, thus the exponent must be equal to zero and it isn't really necessary to put the same base on both sides of the equal.
@thomasmaughan4798 True, but if the instructor requires "Show your work", it is.
@@jamesharmon4994 It seemed ratther arbitrary for someone to subsitute "1" on the right with Log6(0) which produces a 1. So will pretty much any other log base to the zero power. So you CAN choose Log6 which then matches the other side and thus vanishes into the thin air from which it came.
@thomasmaughan4798 actually, it's log6(1) which produces 0. This mistake demonstrates that logs are too complicated for those who "don't know where to start."
I now know where to start, thanks to your relentless teaching and a lot of practice.
Exponential equation? Think Logs!
It worked. :)
Anytime you see a more complicated equation or expression in an exponent try using logs to simplify it.
@@WitchidWitchid Indeed! :)
But you didn't show why anyting raised to zero is one; i.e. why x^0 =1 - I will show it in two ways, the first with numbers to help get the feeling for it, and the other with a in order to "make it mathematically correct as a proof"
16=2^4, 8=2^3, 4=2^2, 2=2^1, 1=2^0, 1/2=2^-1, 1/4=2^-2, 1/8=2^-3, 1/16=2^-4
Beginning with 16, every time you divide by two, the result is that one is subtracted from the exponent, and so your exponent starts at 4 and is reduced to 3, 2, 1, 0, -1, -2 -3, -4 while the number is halved down to 1 and then to 1.2 to 1/4 etc, and the 1 occurs when the exponent is 0.
axaxaxa=a^4, axaxa=a^3. axa=a^2, a=a^1, a/a=a^0, 1/a=a^-1, 1/(axa)=a^-2, 1/(axaxa)=a^-3, 1/(axaxaxa)=a^-4.
Take a4/a =a3 (axaxaxa)/a =(axaxa). Similarly (a^3)/a=a^2. Similarly (a^2)/a=a^1. Similarly (a^1)/a=a^0. Similarly (a^-1)/a=a^-2.
The crucial point is (a^1)/a=a^0. i.e. a^1 = a, and a/a=1 while a^1/a^1 = a^1 x a^-1 = a^(1-1) = a^0. So we have a/a = a^0 and a/a=1
Why not teach your students to think? There is no reason that they shouldn't out right know that 6^0 = 1, so just set the power equal to 0
Yeah, the ‘70s……….we had to actually DO this lol! Ah, the tables in the back of the books. We were allowed to use the NEW scientific calculators in my senior year (1978). They were exorbitantly priced (over $100)! The minimum wage was $2.65 then. It would be like paying over $275 today. My dad was unhappy! I had a friend who used his trusty slide rule. Ah, those were the days!
You know, you talk about loo,ing up logarithmic values in tables. The first year of engineering classes at UMASS/Amherst we had slide rules. And we had to use slide rules in our exams. 😂
Oh, come on. "Anything to the 0 equals 1." So, 3x+5=0 --> x = -5/3.
If you want to "solve it formally," then:
6^(3*x+5) = 1
log[6^(3*x+5)] = log(1) = 0
(3*x+5)*log(6) = 0
3*x+5 = 0
x = -5/3
6^(something) = 1 mean that (something) MUST equal zero.
So, 3x+5 = 0. This changes that exponential equation into a simple linear equation.
3x = -5 → x = -5/3
In this specific case, you don't need logarithm.
I do remember using those logarithm tables... It was quite tedious.
Why do you make simple things so many times so difficult?
at 17:48 we have (3x+5)log6=log1. So. (3x+5)log6=0. So 3x+5=0
Why all those extra steps? The only confuse students.
@TabletClass Math . There is total nonsense in your math steps. You jump all over and make no sense what so ever. On top of all things, you are not able to show bit by bit using logic, how do you ever get there. The final answer may be right but then why any student need to bother to show any steps that makes sense when they could just get the final answer from a calculator and then if need, throw a bunch of numbers and equations that makes no sense and at the end put an equal = FINAL ANSWER. You may be an mathematician but surely are very poorly qualified to teach math since your teaching steps makes NO SENSE WHAT SO EVER. Step by step
means
no jumping all over. It also means that you have to be crystal clear with every number your bring, WHY THAT NUMBER AND FROM WHERE, and must explain why that number and why not another. And whatever number you bring must be brought through logic. There must be formulas or rules that explains and show clearly why that number and why not a different number. You have a huge deficiency in your steps and explanations. This is the
third
math-video-clip we have watched from your channel, and is the third that have missing steps or else steps that makes no sense what so ever. People will go mad if continue to watch this nonsense. We were hoping for the third math video clip to make sense with each step jumping nothing but rather show crystal clear good sense good logic explanation of the entire process step by step, but it did not, and failed again. We totally give up.
Many don’t know where to start: 6^(3x + 5) = 1; x = ?
6^(3x + 5) = 1 = 6^0, 3x + 5 = 0; x = - 5/3
Answer check:
6^(3x + 5) = 6^(- 5 + 5) = 6^0 = 1; Confirmed
Final answer:
x = - 5/3
set the exponent expression = 0 and solve for x --> 3x + 5 = 0 b/c base^0 = 1
3x + 5 = 0
3x - 5 = -5
x = -5/3
6³ˣ⁺⁵ = 1
My solution:
Therefore 6³ˣ. 6⁵ = 1
Therefore 3x = -5
Therefore x = -5/3
Now I need to see if I can get there using logs. Thanks for the tips.
On Indices any number raised to zero gives 1 therefore the power on the right is zero meaning equating 3x+5=0 solving gives X=-5/3 .The explanation may confuse learners.
Another method would be to take the log of both sides and one would get (3x + 5)log6 = log1. One would then get 3x + 5 = log1/log6. The next step is 3x = log1/log6 - 5. The solution
would be x = (log1/log6 - 5)/3.
24 minutes…😢
6^(3x+5)=1 which means 3x+5=0 ergo 3x=-5 ergo 3x--1.66 Simple highschool stuff. No need for logarithms.
I got -1.66 why is that wrong? Should I have left my answer in a fraction? I know it means nothing, but, this is fun. I've had much higher calculus than any of this, but I apparently have a lot to learn?
For those wondering why he used logarithms, I am fairly sure that that was the whole purpose of the lesson.
From above:
"How to solve an exponential equation using logarithms."
6^(3x+5)=1 so taking logs of both sides and dealing with the exponent we get (3x+5)*log(6)=log(1), and thus 3x+5=log(1)/log(6), but log(1)=0 (for logarithms of any base) so 3x+5=0, so x=-5/3.
For that equation to be true, 3x+5 must equal 0, so x=-5/3
TabletClass Math doesn't know where to start. He should have picked a number other than 1 which can be converted to 6^0 if his intent was to teach logarithmic solutions to equations.
How many used Log tables in the back of your Algebra 2 book and use Trig tables??
i am put off by the pure average wording an d how ut is disinteresting kind of jusut before the go to statements to b e honest kind of based of what my name means and judgement and interest for maths regardless pf knowing it should be set to 0 meaning the exponential expression if it is ok with you to call it t hat where i would like to call it an equatiom for the top part but is an equation overall meaning 6^3x +5=1
Really simple!! We know that any number to power 0 = 1, So, 3x + 5 = 0 !! Therefore 3x = -5 , and so x = -5/3
6^(3x+5) = 1 = 6^⁰
Therefore 3x+5 = 0. Exponent values are equal
3x= -5
X = - 5/3
QED
Any number to the zero power is 1. Set 3x+5 =0 and solve for x. No need for logarithms or any of that for this problem.
For any y, y^0 = 1
Therefore, 3x + 5 = 0
3x = -5
x = -5/3
Where is the challenge? I did this in my head.
please say right lets dive into it so it sounds more professionial and to the point period and for also how it should sound instead of just t alking which sort of isn't really interesting oj its own before the go to statements
,... ,,,.,,.
This is wrong, but I'll give arrogant conjecture and say -15/3. Laugh as hard as you wish.
seriously if someone doesn't understand what an exponent is, this video is way too advanced for them. You didn't need to spend 5 minutes explaining what an exponent is in going into logs.
more known as Shiva
unless misaken not the actual god the Ramanga mathmatician known as Shive as welll where it is Ramanag Shiva but i feel a bit wrong where it was and am will to take critisome on the 2nd name
Most simple way is to make the power of 6 zero, so because 3x +5 to be zero needs 3x to be equal to -5 so x should be -5/3
Simpler: If (3x + 5)log 6 = log 1 = 0,
then since log 6 not= 0, 3x + 5 = 0, and x ≈ -5/3.
6^(3x+5)=1=6^0
compare indices,
3x+5=0
x= -(5/3). It took me about 3 secs.
What application do you use for your presentation. What application do you write abd draw on please?
Interesting tooic...math...but you talk too much.other math vloggers are better
At first it looks like a hard and tricky problem due to the x variable in the exponent. But it is really a very easy question.
Just start taking logs from both sides, which eliminates the 6, and solve the equation.
Maybe you should have had this equation equal something other than 1, since knowing that anything to the zero power =1. So if 3x+5 is zero, then it's really easy and doesn't require logs.
Anyway, thanks for the instruction on logs. They always kinda perplexed me, and you're helping me to understand.
ok hearing you say bacon and egg makes me immediately want to click off the video. Please don't try to be cute.
A much easier way. 6 to the zero power is 1. As such 3x + 5 = 1. Took me less than 3 seconds to solve.
for thinking and facts where it is or isn't and wht was meant by each individuak term ,,. if i do i do for those who are india and think i am lying if i am i am period
Any number not zero, to the power of zero equal 1. Thus 3x+5 = 0, or x = -5/3.
Es más fácil esta solución
6^(3x+5)=6^0
3x+5=0
3x=-5
x=-5/3
it really should be more interesting for the go go go comments actually i have to say this back instead ,
6 to the power of zero equals 1. Hence, 3x+5=0; this yields x=-5/3
It is Not the Answer that you are learning here. Its the Method so can apply to every others. In Maths Methods solves.
I was in my final two years of High School (5th and 6th Forms or Years 11 and 12) here in Australia when scientific calculators became available, but they were equivalent to about a week's pay or more so not many people could afford them. Also, we were not allowed to use any calculator in exams. Calculators were only used to double check our answers and even then, you could not trust calculators to be correct at anything other than basic addition and subtraction.
Greetings. I can remember those days very well.
@@devonwilson5776 I also remember using a slide rule and it cost a whole lot less than even a basic calculator.
I was a poor graduate student armed only with a slide rule when scientific calculators first became available. I was the only student not using a scientific calculator because I also had a wife and kids to feed. Scientific calculators should never be banned from any exam when they are not banned from the job. Instead, the student should be required to justify every step taken to arrive at any answer obtained from a scientific calculator. For example, suppose my boss handed me this new equation he just discovered, (2.3)^(2x)-4*(2.3)^x+4=0, and ordered me to have a value for x by tomorrow. And by the way, scientific calculators are not allowed. I know several ways to solve this problem, but I can't think of a useful way to get a value for x without a scientific calculator. For example, substitute y=(2.3)^x into the given equation to get y^2-4y+4=0, which has roots y=[4±√(16-16)]/2=2, so (2.3)^x=2. Of course I could then take the natural log of this equation to get x*ln(2.3)=ln(2) and solve for x=ln(2)/ln(2.3), but without a scientific calculator I couldn't provide a numerical value for x without consulting my old "CRC Standard Mathematical Tables and Formulae" book.
If you can't trust those calculators, then they're useless. What you couldn't trust is the user, if he/she didn't understand how the device operates and its limitations.
I think the real problems are that most early ones had no support for extended precision, floating point (mantissa & exponent), precedence rules, parentheses, and memory store/recall. They also lacked many common functions, such as roots, powers, exponentiation, logarithms, circular & hyperbolic functions along with their inverses. That means you still need to make sure you don't exceed the range of the calculator, use look-up tables for the common functions, be wary of precision errors introduced, and have scratch paper on hand to record intermediate results that will be hand-entered later on. That's where errors get introduced into the calculations.
My sister got a TI SR10. It has built-in [x²], [√x], and [1/x] functions, but no parentheses or memory store/recall. It can display the result in mantissa with exponent format. Neatest thing since sliced bread for engineers and scientists.
Multiply both sides with 6^-5 => 6^3x = 6^-5 => 3x=-5 .............
This is linear equation, if someone cannot solve it they are probably humanist :)
That is simple
Why make it complicate with logs?
This is a sixth grade problem
whenever an answer to a question is approx. i sign off. that's not an answer, that's a guess.
So many people got it right although not knowing where to start. They must be stupid
your method is way too complicated. Nothing i can add. everyone has already said it
There’s another way and is elevate both terms of the equation to 0 And so forth and so on.
Too much explaining. Just get to solving the problem.
Hi John you are an excellent teacher. I am having a lot stuff cleared up
If we assume that any number that has 0 as an exponent is equal to 1, then 3x+ 5 = 0
Franchement c'est la méthode la plus chronophage que j'ai vue :)
U teaches this very confusing, u r all over the place, was getting it then lose it
only 3x+5 = 0 will be a "real" solution to make that equation true.
I went straight to the 3x+5=0 because any number x zero = 1
wow, i actually deduced that the exponent had to be 0 for the equation to = 1. dang, sometines i amaze myself. all in fun.
thanks for a great lesson, & actual reassurance.
i went to HS in the mid-late 60s. if you got caught with a calc, YOU FAILED, sometimes the test or even the semister. DONE !!!
All good thing comes in three's
Greetings. I remember those days, for me 70's, in primary school, absolutely no calculator allowed. The good old days.
That’s because calculators are for students who understand and therefore know how to do the mathematical calculations without using a calculator. What if the calculator broke down and we didn’t know how to arrive at the answer way the instructor is doing here? It’s easy to hit 5 and then log10 on the calculator. But what if the calculator pegs out and break right at that moment and we didn’t know how to do the problem by hand because we didn’t know and understand how to do the problem by head and hand? Calculators are excellent for doing tests and even during class, but only after a subject has been thoroughly understood and learned by the head and done by the hand repetitiously until committed to both memory and understanding. Does no good to have students hit and peck 2 + 2 on the calculator when they don’t understand and know how to do addition through understanding and repetitive practice. What happens if the instructor gives a set of problems to be done within a reasonable time frame but says, “absolutely no calculators!” Then what! Calculators are for those who understand and know the work, they’re not for the uninitiated! The uninitiated will never understand and learn the subject matter properly and thoroughly by simply punching in log100 with a calculator… they need to first understand and learn why the calculator gives the answer 2! For example, why would I add 0.3456889731237 + 35.903578134556 = x by head and hand instead of using a calculator when I already understand and know how to add decimals by head and hand? It would consume way to much time and be completely insane for the teacher to have me to write that problem out by head and hand instead of using a calculator when the teacher knows that I’ve proven and therefore understand and know how to do the problem by head and hand! But for the uninitiated student they must necessarily begin by first understanding and knowing and doing the problem of adding decimals before they move on to calculators. Calculators are horrible for students who don’t first understand know how to do math by head and hand through practice, practice, practice! So… what do I do with 30.4567893400586 - 7.8967855422106 if I didn’t know how to subtract decimals by head and hand but knew how to peck out the answer on a calculator but all the calculators in the world were broke….?
Back in the good old days you used a slide rule. I used tables and a slide rule in 1955.
Yes, but in his 2 problems you can do them easily, right?
6^(3*x+5) = 1
Take log base 6 of both sides:
3*x+5 = 0
x = -5/3 ≈ -1.666667
4^y = 10
y*log(4) = log(10)
y = 1/2/log(2)
We remember log(2) ≈ 0.30103, so
y ≈ 1/0.60206 ≈ 1.660964 .
You dont need 24 minutes ( ! ) to explain this stupid problem
Parece fácil :
1=6^0 assim 3X+5=0 X=-5/3 e pronto !!! 🤷🇧🇷
You keep digressing too much. It's a poor habit.
I'll solve this equation in three different ways, and one should know all three of them.
(1) 6^(3x+5)=6^(3x)*6^5. Divide both sides by 6^5: 6^(3x)=6^(-5). Equate powers of 6: 3x=-5 so x=-5/3
(2) Any number raised to the zeroth power is 1, so 6^0=6^(3x+5)=1 yields 3x+5=0 and x=-5/3
(3) The natural log of 1 is 0, so take the natural log of both sides: (3x+5)ln(6)=ln(1)=0, so (3x+5)=0/ln(6)=0 yields 3x+5=0 and x=-5/3
Unncessary a long and boring solution can be solved in 3 or 4 steps।
I would just set 3x+5 as t and solve t first and then substitute it.
It's easy, exponential and logarithmic
Why log? Didn't have to make it so difficult.
could you use x = -(5/3) to solve the problem?🤔
Is someone able to build a solution in Excel/Google? I'm having a hard time solving for a complex exponent in Google/Excel. This is based on a financial formula.
A=P*((1+r/n)^(n*t))+x
Solving this for (t)
(A-x)/P =(1+r/n)^(n*t)
At this point, I think I need to use a log function to get the exponent out, but if I capture (t) in "log(1+r/n,n*t)", I'm really not sure what to do to get (t) out of the function.
Please help!
generally for equations a^(bx + c) = d, you can take ln of both sides so ln(a^(bx+c)) = ln(d), then use the logarithm power rule to bring bx + c to the outside, yielding, (bx + c)(ln(a)) = ln(d), divide both sides by ln(a), bx + c = ln(a) / ln(d), by logarithm rules ln(a) / ln(d) = log base d of a which i will write as log_d(a), subtract by c to get bx = log_d(a) - c, divide by b for the final answer of x = (log_d(a) - c) / b. if we plug in the numbers, a = 6, b = 3, c = 5, d = 1, x = (log_1(6) - 5) / 3, log base 1 of all positive real numbers (including 6) is 0, so x = -5/3
Even one sec mentally is too much for this question.
Sir you are underrated you deserve more views and subs
Views is what count$
JEEZ NO WONDER STUDENTS HATE MATH 25 MINUTES FOR THIS
Don't forget to mention the slide rule!
We used slide rules when I was a junior, but we could use calculators as seniors. The school bought them and we could borrow one. The first Calc I used was the HP which used polish infix notation. It did everything but graph. Switched to TI as undergrad. Returned 20 years later for advanced degree, and we used a nice--TI graphing calculator.
But for advanced Calc, discrete math, etc--we used laptops and Maple and Mathematica (now Wolfram). Once mastering these tools, you just couldn't miss homework problems.
The best thing about these tools, was the ability to try all sorts of approaches. If an approach failed, you learned something. Playing around and thinking really expands your depth of knowledge.
One thing to remember is there are no timed tests in the workforce-- only deliverable dates. Sure, some
things needed quickly reviewing, but some problems took months, so they were always in the background, to be attacked as insight lit up connected neurons :-)
Why is it that most "aha" moments occur in the shower, working out or in dreams?? 😮 I think the subconscious is handling a lot of work so the conscious mind can function on the immediate tasks at hand.
Man, I got pretty metaphysical here.
This can be solved using derivatives, right?