I'm your age... My Brother was E. ENG and my Dad was in Chem. Eng. I'm going back and (re)- learning. This guy has great vids. You're absolutely right.
Mr H is just FANTASTIC. He has sent me back to the 70s and I would love to keep pace with him. He makes Mathematics FUN. Straight to the point but we would sometimes like to have some tit bits from him and to introduce some books for us to acquire, without promoting others too much if he has some books or Pamphlets on Logarithms and Indices and Trig and Calculus, some of us would love to acquire them. Thanks Sir. We love you Mr H.
Answer x = 8. 359 A different approach. Log 2 (x) + Log 3 (x) = 5 This means: 2 raised to a number = x added to 3 raised to a number also =x and the sum of these = 5. let p = the number 2 is raised to; let n= the number 3 is raised to; Hence p + n =5, and hence, n= 5-p, So: 2^p = x, and 3^ 5-p = x Hence, 2^p = 3^5-p (since both = x, transitive property, if x=y and r=y, then x=r) log (2^p) = log (3^5-p) ( log both sides) p log 2 = (5-p) log 3 (rules of log) p = (5-p) log 3/log 2 p = (5-p) 1.58696250072 p = 7.92481250361 - 1.58696250072 p p + 1.5869p = 7.9248.. 2.58696 p = 7.92848 p = 7.9248/ 2.58696 p = 3.0633658 recall x = 2^p and we are solving for x , hence, x = 2^ 3.0633658 = 8.359 answer Verifying if 8.359 is correct note that since p = 3.0633658 then 5-p = 5- 3.0633658 = 1.9366342 and recall x also = 3^ 5-p = 3^ 1.9366242 which also = 8.359 p
... Good day to you, One short remark regarding approximation, maybe it would be better to approximate X at the very end of the algebraic calculation, instead of approximating 1/log2 + 1/log3 (5.42) and then again with 10^(5/5.42) ... obtaining less accuracy? Thank you for your math efforts ... happy weekend, Jan-W
Thank you for pointing that out. I thought about keeping the exponent as 1/log2 + 1/log3 as well. I calculated the final value with 1/log2 + 1/log3 as the exponent and also 5.42 before making the video. Difference between two was very minimal.
If calculators aren't allowed and by looking at the second step, then we can find the LCD, which is log(2)log(3). Multiply everything by that LCD, which is (log(3)+log(2))log(x)=5log(2)log(3). Divide both sides by log(6) (from log(3)+log(2)), which is log(x)=5log(2)log(3)/log(6). There are 4 acceptable answers for solving for x, which are x=32^(log_6(3)) or 3^(log_6(32)) and 243^(log_6(2)) or 2^(log_6(243)).
If calculators aren't allowed, I hope that logarithm tables are ! In this case, the best way is to transform the initial equation in order to get an expression with any logarithms you may find in the available tables. If no table is provided (or allowed) then your last hope is to graphically hint for the solution on a sheet of logarithmic millimeter graph paper.
I don't really spend much time with math any more but I'm still interested in the subject. One of the things you can do to work problems like this, if you aren't allowed a calculator, is to spend a little time making up a mental toolbox of powers and roots, and logarithms of primes. It takes a little time to do the memorization, and you'll have to keep reviewing occasionally to refresh your memory, but it makes this much easier.
I'm not good at that. The only thing I took away from here, for the moment, is the transformation of the exponent on an old basis into an quotient representing the exponent on a new basis. But, at least, that worked ... 😊
This exact example, unlikely. Logs in general, they have plenty of applications. Some examples I can contribute: 1. Cooling and heating, and predicting the time for heat to transfer. To invert the exponential decay function that tells temperature vs time, you use logs to do so. 2. Decibel and octave scales, that translate frequency and intensity into a form people can understand as pitch on the music scale and the decibel scale for sound level. Human hearing is logarithmic in nature. 3. Electrical timer circuits. Electrical timer circuits, use an exponential decaying signal to keep track of time. The resistor-capacitor charge/discharge behavior is an exponential decay. To invert it and determine time, you'll use a logarithm function.
It is the base of the logarithm used to unify the bases. In this example decimal log was used, so both sides were raised as exponents of base 10 in order to get x on the left side. You could use ln - natural logarithm to replace log2 and log5, and in that case you would use e instead of 10 here.
Depends on context, and on the rules of your calculator. The short answer is that it is most common for log(x) to mean log base 10 of x by default. Most scientific calculators and Google Calculator use this convention. In pure math, the notation is also used for log base e. Ln(x) would mean the real numbered logarithm function, while log(z) would be the complex log. Log base e is preferred in pure math, because its calculus is the most elegant. Wolfram Alpha uses this meaning of the log function, so you have to rename it to ln(x) when copying it to other programs. Less common is for log without a specified base, to mean log base 2. Commonly preferred in computer science applications, though I'm not aware of any software meant for the general public that uses this convention.
... LOG(X) means LOG(base 10)(X), so suppose LOG(X) = LOG(base 10)(X) = 5 , this means that X = 10^5 ... LOG(X) doesn't show its base 10 ... hope this helps you ... Jan-W
the expression "Log" without an explicit base is defined as Log to the base of 10. In this example he just took the second equation where he can substitute the log of b to the base of a on the one hand and the division of log of b to the base of c and the log of a to the base of c on the other hand. C is not defined so he could use any base. He chose to use the base of 10 though he said he would prefer the log to the base of e (ln).
To be honest: if all logarithms are brought down to the basis of e, do you think there is a way to solve any of these equations without a calculator???
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brilliant! i loved logarithmic equations in school, but we never learned to solve an equation with different bases. thank you!
I’m going back to school at 42 for electrical engineering and your videos have helped me immensely. Thank you!
Good for you. Congratulations!
Very admirable!
Same! Best of luck!
Bless You. I walked into LSU at 30 years old (Chemical Engineering) Most of my classmates thought I was the Professor.
I'm your age... My Brother was E. ENG and my Dad was in Chem. Eng. I'm going back and (re)- learning. This guy has great vids. You're absolutely right.
Mr H is just FANTASTIC. He has sent me back to the 70s and I would love to keep pace with him. He makes Mathematics FUN. Straight to the point but we would sometimes like to have some tit bits from him and to introduce some books for us to acquire, without promoting others too much if he has some books or Pamphlets on Logarithms and Indices and Trig and Calculus, some of us would love to acquire them. Thanks Sir. We love you Mr H.
I am always blown away when I learn more about logarithms
I am going back to the benches at 68 to complete my MD degree.
I am also greatful for these scirnce videos.
Congrats! I’m about start a second bachelors in physics after 20 years.
@@ToTheWolvesall the best 👍
All the best 👍
Love hearing that ❤
@@88kgs ty
Making maths beautiful.
Thank you sir.
🙏
Thank you for the nice comment.
Perfect explanation!
Very useful professor 👍
Answer x = 8. 359
A different approach.
Log 2 (x) + Log 3 (x) = 5
This means: 2 raised to a number = x added to 3 raised to a number also =x
and the sum of these = 5.
let p = the number 2 is raised to; let n= the number 3 is raised to;
Hence p + n =5, and hence, n= 5-p, So:
2^p = x, and
3^ 5-p = x
Hence, 2^p = 3^5-p (since both = x, transitive property, if x=y and r=y, then x=r)
log (2^p) = log (3^5-p) ( log both sides)
p log 2 = (5-p) log 3 (rules of log)
p = (5-p) log 3/log 2
p = (5-p) 1.58696250072
p = 7.92481250361 - 1.58696250072 p
p + 1.5869p = 7.9248..
2.58696 p = 7.92848
p = 7.9248/ 2.58696
p = 3.0633658
recall x = 2^p and we are solving for x ,
hence, x = 2^ 3.0633658 = 8.359 answer
Verifying if 8.359 is correct
note that since p = 3.0633658 then 5-p = 5- 3.0633658 = 1.9366342
and recall x also = 3^ 5-p = 3^ 1.9366242 which also = 8.359
p
Yours is much easier to follow. ❤
not a pure answer, ew
... Good day to you, One short remark regarding approximation, maybe it would be better to approximate X at the very end of the algebraic calculation, instead of approximating 1/log2 + 1/log3 (5.42) and then again with 10^(5/5.42) ... obtaining less accuracy? Thank you for your math efforts ... happy weekend, Jan-W
Thank you for pointing that out.
I thought about keeping the exponent as 1/log2 + 1/log3 as well.
I calculated the final value with 1/log2 + 1/log3 as the exponent and also 5.42 before making the video.
Difference between two was very minimal.
@@mrhtutoring I made that remark because of my technical background (lol) and thanking you for your clear explanation ... take care, Jan-W
If calculators aren't allowed and by looking at the second step, then we can find the LCD, which is log(2)log(3). Multiply everything by that LCD, which is (log(3)+log(2))log(x)=5log(2)log(3). Divide both sides by log(6) (from log(3)+log(2)), which is log(x)=5log(2)log(3)/log(6). There are 4 acceptable answers for solving for x, which are x=32^(log_6(3)) or 3^(log_6(32)) and 243^(log_6(2)) or 2^(log_6(243)).
If calculators aren't allowed, I hope that logarithm tables are ! In this case, the best way is to transform the initial equation in order to get an expression with any logarithms you may find in the available tables.
If no table is provided (or allowed) then your last hope is to graphically hint for the solution on a sheet of logarithmic millimeter graph paper.
I don't really spend much time with math any more but I'm still interested in the subject. One of the things you can do to work problems like this, if you aren't allowed a calculator, is to spend a little time making up a mental toolbox of powers and roots, and logarithms of primes. It takes a little time to do the memorization, and you'll have to keep reviewing occasionally to refresh your memory, but it makes this much easier.
At the beginning of the video, I thought the solution would go in the direction of the LCD, and we would leave the calculator in the drawer.
Thank you sir, please keep it up
Good keep up.
Very clear! You are a great teacher! Compliments!
Splendid.
Good stuff. Thank you.
Thanks for the vedio
Dude thank you so much 😊 your the best man!
Yoooo thank you I got an A- after failing 3 tests back to back and studying 4 hrs each thank you bro
Good job!
Btw would you consider making a video on circles, ellipses, parabolas, and hyperparabolas.
I will try~
Thanks!
Great job🎉
Proffesor, how come you prefer ln to log?
2 letters vs 3 letters
Lol!
@mrhtutoring 😊😊
Thank you for the video.
Magnificent
Can you do derivatives and intagration and sigma rule ❤
Am asking for the same here 😊.
I'm not good at that. The only thing I took away from here, for the moment, is the transformation of the exponent on an old basis into an quotient representing the exponent on a new basis. But, at least, that worked ... 😊
With the exact values, it would round up to 3.373
x = e^(5 ln 2 ln 3 / ln 6)
looks cleanest
👍❤🙏🙏🙏
5.4178313691767469521668869515108 and 8.3658336590275351036590017462244 to be more exact. I had to do the math. :D
🎯🏆🏁
One simple question...when do we need to use this logarithm equation on our daily life or real life situation?
It depends on your situation. You probably experience it and don't know it. The decibel scale and the Richter scale are both logarithmic.
Logarithmic equations are used a ton for scientific calculations. One example is chemistry ( which im studying)
pH (concentration of hydrogen cations) is also logarithmic..
This exact example, unlikely.
Logs in general, they have plenty of applications. Some examples I can contribute:
1. Cooling and heating, and predicting the time for heat to transfer. To invert the exponential decay function that tells temperature vs time, you use logs to do so.
2. Decibel and octave scales, that translate frequency and intensity into a form people can understand as pitch on the music scale and the decibel scale for sound level. Human hearing is logarithmic in nature.
3. Electrical timer circuits. Electrical timer circuits, use an exponential decaying signal to keep track of time. The resistor-capacitor charge/discharge behavior is an exponential decay. To invert it and determine time, you'll use a logarithm function.
@@carultchthanks for the answer 🙏
Sadly in my Country we dont plug the log into the calculator :(
wait so will the value always be 10? ( here, 3:12 ) or is there a condition??
It is the base of the logarithm used to unify the bases. In this example decimal log was used, so both sides were raised as exponents of base 10 in order to get x on the left side. You could use ln - natural logarithm to replace log2 and log5, and in that case you would use e instead of 10 here.
Depends on context, and on the rules of your calculator. The short answer is that it is most common for log(x) to mean log base 10 of x by default. Most scientific calculators and Google Calculator use this convention.
In pure math, the notation is also used for log base e. Ln(x) would mean the real numbered logarithm function, while log(z) would be the complex log. Log base e is preferred in pure math, because its calculus is the most elegant. Wolfram Alpha uses this meaning of the log function, so you have to rename it to ln(x) when copying it to other programs.
Less common is for log without a specified base, to mean log base 2. Commonly preferred in computer science applications, though I'm not aware of any software meant for the general public that uses this convention.
I like You matter
And that's why I hated Maths, keeping track of all the different principles
Can someone tell me where the 10 comes from😭
... LOG(X) means LOG(base 10)(X), so suppose LOG(X) = LOG(base 10)(X) = 5 , this means that X = 10^5 ... LOG(X) doesn't show its base 10 ... hope this helps you ... Jan-W
@@jan-willemreens9010 thanks alot
the expression "Log" without an explicit base is defined as Log to the base of 10.
In this example he just took the second equation where he can substitute the log of b to the base of a on the one hand and the division of log of b to the base of c and the log of a to the base of c on the other hand.
C is not defined so he could use any base. He chose to use the base of 10 though he said he would prefer the log to the base of e (ln).
Logx is in base 10
Log (no base) is assumed to be ten
I got sad when he solved the rest of the problem with a calculator.
To be honest: if all logarithms are brought down to the basis of e, do you think there is a way to solve any of these equations without a calculator???
stop rounding answers!
Can you not round a log answer?
Great ❤
I believe we should use 5.417831369177, instead of just 5.42. Because we need the final answer to be 3 sig fig.