I like your method way of turning a single thought in base to a power to a product of two entities in taking the logarithm on both sides mathematical manipulation. Someone in the comments clarified why any base to an "even integer power" of case 3 to add as left and right side of the given equation mathematics manipulation to cleverly find the method 3 in the video solution without thinking extra about the "even powers of even or odd integer bases" trick by: Start with the given: x^(x + 3) = 1 Instead of taking log() in both sides instead multiply by "x" on both sides to get the equation x^(x + 4) = x. You can then use that newly formed equivalent equation for negative x where log(negative x) are not defined in real numbers that include negative numbers. ... Running through negative x values other than (-3)^(1) = -3 result found in second method of the video x + 3 = 0, the newly formed x^(x + 4) = x newly equivalent equation shows (-1)^(-1 + 4) = (-1)^3 = (-1) result in the substitution of x = -1 in case (-1)^(-1 + 3) not equal to (-1) but (+1) was mote confusing (or the lost train of mental thoughts of any even or odd integer to an even power is a positive integer result).
@@Sglagoomio It's not the 1 that turns into |x| ln(|x|^(x+3))=ln(1) (x+3)ln|x|=0 You might ask why I introduced the | | here. Compare it with solving an even power equation like x^2=16 with logs
Solution to all 3 questions Q1. Case 1: base is 1 x + 2 = 1 ⇒ x = -1 Case 2: exponent is 0 and base is non-zero exponent = x + 3 = 0 ⇒ x = -3 base = -3 + 2 = -1 ≠ 0 Case 3: base is -1 and exponent is even base = x + 2 = -1 ⇒ x = -3 exponent = -3 + 3 = 0 , which is even Therefore, x = -1 or -3 Q.2 base = x + 2 = -1 ⇒ x = -3 exponent = -3 + 3 = 0 , which is NOT odd Therefore, no real root. Q.3 base = x + 2 = 0 ⇒ x = -2 exponent = -2 + 3 = 1 ≠ 0 Therefore, x = -2
Q:1 X=-1(base=1) X=-3(0 exponent) Base=-1 would give us -3 as solution too so there are only 2 solutions Q:2 A exponent can only be -1 if the base is -1 and it's expoenent is odd so we would get x=-3 for base 1 but that would give us exponenet of 0 which is even so there is no solution Q:3 A exponenet can be equal to 0 only if the base is 0 so we would get x=-2
For some reason I did this in my head by using the natural log and stuff, but I did manage to get -3 and 1. I then looked at it again and knew I didn’t need that at all. At least I did it in my head :P
That way was close to proving case 3 anything to an even power is positive. So if x^(x + 3) = 1 was figured out in cases 1 as +1 to any power odd or even (since +1 is an odd integer then 1 + 3 = 4 is an even integer) ... and 2 of odd integers -3 and 1 raised to an even power themselves, respectively. We are still missing one other odd integer raised to an even power from the negative integers after the x + 3 = 0 solution giving x = -3 of (-1)^(2) one of case 3! ... Then you were starting to show case 3 proof by multiplying both equation sides by x to get x^(x + 3) times x = 1 times x ... Or ... x^(x + 4) = x that would have shown his case 3 result (-1)^(-1 + 4) = (-1) parallel to saying any integer positive or negative to an even integer power is a positive integer result! Good try! 👍
An other solution using log function we use the absolut value to have the function defind in R (X+3) log(|X|) = 0 => X+3 = 0 => x = -3 or X = 1 or X = -1 (because X+3 is even)
That’s how I was thinking but forgot about the absolute value so just got positive 1. Looks right but the reasoning behind x=1 and x=-1 as solutions is because it makes ln|x| equal zero not because x+3 is even. Well I guess it’s true both ways but it definitely makes the natural log function zero.
Just a quick question to the wonderful admin or anyone who reads this. If I start learning maths from grade 10(other than previous grades), will I know enough to attempt engineering entrance exams? Like the examples above and so on? Thanks
Hellooooo. Hadnt u uploaded a video of P(x) ^ Q(x) = 1 back in the past? Where solutions where 2 3 4 5 6 7 and wolfram alpha was missing 2 of them? Pls, if i am right, could you link it? I cant find it :(
solution to Q1: case 1: 1 to the power of any number is 1 so x+2 = 1 x = -2+1 = -1 case 2: any number to the zeroth power is 1, so x+3=0 therefore, x = -3 case 3: (-1)^even is 1, so lets put x+2=-1 so, x = -1-2 = -3 wait...no nevermind let's keep going the exponent will be -3+3=0 so, (-1)^0 = 1? TRUE since the solution to case 2 and 3 are the same, I will only write it once so finally, x = (-1,-3) solution to Q2: (-1)^odd = -1, so we'll let x+2 = -1 => x = -1-2 = -3 putting that into the original, (-1)^(-3+3) = (-1)^0 = -1? FALSE so no real solution solution to Q3: 0 to any positive real number is 0 so, assuming x+2 = 0, x=-2 putting that into the original, 0^(-2+3) = 0^1 = 0? TRUE so, x=-2
I just found a more straightaway solution to Q1 using logarithms take ln on both sides and bring the power to the front to get (x+3)*ln(x+2) = 0 x+3 = 0 x = -3 ln(x+2) = 0 x+2 = 1 x = -2+1 = -1 so again, x=(-1,-3) now I found a flaw in this if you do this, you are limiting yourself from ever getting the answer from case 3 however, I found a solution and it is that you can let the argument be -1 and if x+3 is even, you can bring the power up to the exponent of -1 to get 1 inside the ln and finally get ln(1) = 0 we are just lucky to have case 2 and 3 yielding the same answer, otherwise we'd have another solution that we would've missed
@@Brid727 ..ln(x+2)=0 ... ln(anything)=0 will be undefined, ur asking what exponent of '0' makes 'x+3' (since u set it equal to zero), since there exist no real solution its undefined 0^anything=0, 0^0 is undetermined/undefined Basicly e^zero=0 (wich has no real solution) however.. ln(x+3)=ln(x)ln(3), ln(3) is close to ln(e)=1[e^1=e] , 1.09... -> 1.09 * X≈0, for practical reasons we can assume the limit x as zero, but as a logarithm ln(zero) is undefined ..in case x would be infinitassible small but not zero...
@@Patrik6920 ln(1) = 0 laughing in the corner 😂 also please read what I wrote above and plug x=-1 in and if you have the slightest bit of knowledge in logarithms you'd know that ln(1)=0 I think you are confusing what I wrote with ln(0) which is clearly NOT what you get for plugging x=-1
I have a question. Is there anyway to solve this without logic? I imagine you could use the Lambert W function in some way, but I don't know how. If any commenter knows, feel free to educate me.
@@acetylsalicylicacid You do get the third answer, actually. Log is only defined for non-negative arguments, now since x^(x+3) is 1, we know that it's positive, but we do not know whether or not x itself is positive or negative. So when you pull the exponent out front you need to include absolute value signs, so we actually get (x+3)ln|x| = 0, and ln|x|=0 gives us both 1 and -1.
X=-1 also works since it makes the base 1 2nd assignment requires (-1)^odd but for base to be -1 x would be -3 which would make exponent even and therefore not -1.
You won't get the L'Hopitals limit in base "x" to the "x + 3" when x = 0 or x + 3 = 0. In those cases the (0)^(3) is 0. The (-3)^(0) will equal "1" he shows as case #2. .... The one your brain was saying as a third solution was instead saying his case #3. We forget any integer base to an even power is a positive number result.
Can you teach me how can I sove this problem, please? sqr(a)+sqr(ab)+sqr(abc)=12 sqr(b)+sqr(bc)+sqr(abc)=21 sqr(c)+sqr(ac)+sqr(abc)=30 Find: (a^2 + b^2 + c^2)
If you want to _avoid_ the debate, then you can't just come out and say that the base can't be 0, because then you get me here in the comments debating it. Say that _maybe_ the base can be 0, we'll come back to it if necessary; solve x+3=0, find that the base is _not_ 0, and now you don't have to deal with it. Everybody's happy, the only real solutions are definitely 1, −1, and −3.
Check:
x=1
1^(1+3)=1^4=1 ✅
x=-3
(-3)^(-3+3)=(-3)^0=1 ✅
x=-1
(-1)^(-1+3)=(-1)^2=1 ✅
@@vovkkawhat
Bruh what are u on@@vovkka
1 is also the answer @bprpmathbasics
Now solve it in the complex world 😁
Honestly this is what I expected to happen.
@@jemandanderes7075you shouldn't, this is the math basics channel
@@jemandanderes7075Same
Take ln on both sides
(x+3)ln|x|=0
x=-3 or x=+-1
That's what I thought too
How do you turn the = 1 into a |x|?
@@Sglagoomio The complex logarithm is logz = log|z| + i(argz + 2*pi*n). He's just taking the real part.
I like your method way of turning a single thought in base to a power to a product of two entities in taking the logarithm on both sides mathematical manipulation. Someone in the comments clarified why any base to an "even integer power" of case 3 to add as left and right side of the given equation mathematics manipulation to cleverly find the method 3 in the video solution without thinking extra about the "even powers of even or odd integer bases" trick by:
Start with the given: x^(x + 3) = 1
Instead of taking log() in both sides instead multiply by "x" on both sides to get the equation x^(x + 4) = x. You can then use that newly formed equivalent equation for negative x where log(negative x) are not defined in real numbers that include negative numbers. ...
Running through negative x values other than (-3)^(1) = -3 result found in second method of the video x + 3 = 0, the newly formed x^(x + 4) = x newly equivalent equation shows (-1)^(-1 + 4) = (-1)^3 = (-1) result in the substitution of x = -1 in case (-1)^(-1 + 3) not equal to (-1) but (+1) was mote confusing (or the lost train of mental thoughts of any even or odd integer to an even power is a positive integer result).
@@Sglagoomio
It's not the 1 that turns into |x|
ln(|x|^(x+3))=ln(1)
(x+3)ln|x|=0
You might ask why I introduced the | | here. Compare it with solving an even power equation like x^2=16 with logs
Solution to all 3 questions
Q1.
Case 1: base is 1
x + 2 = 1
⇒ x = -1
Case 2: exponent is 0 and base is non-zero
exponent = x + 3 = 0
⇒ x = -3
base = -3 + 2 = -1 ≠ 0
Case 3: base is -1 and exponent is even
base = x + 2 = -1
⇒ x = -3
exponent = -3 + 3 = 0
, which is even
Therefore, x = -1 or -3
Q.2
base = x + 2 = -1
⇒ x = -3
exponent = -3 + 3 = 0
, which is NOT odd
Therefore, no real root.
Q.3
base = x + 2 = 0
⇒ x = -2
exponent = -2 + 3 = 1 ≠ 0
Therefore, x = -2
For qn 3 you have to state that the power is positive, not just non-zero. Because 0^neg is undefined.
But otherwise very nice solutions.
@@Ninja20704 I forgot that 1/0 etc. is undefined. Thank you.
Nice 👍
You say pause the vid as if I understand anything, thank you
Me (who tried to use Lambert W funcion) 🤡
Q:1
X=-1(base=1)
X=-3(0 exponent)
Base=-1 would give us -3 as solution too so there are only 2 solutions
Q:2
A exponent can only be -1 if the base is -1 and it's expoenent is odd so we would get x=-3 for base 1 but that would give us exponenet of 0 which is even so there is no solution
Q:3
A exponenet can be equal to 0 only if the base is 0 so we would get x=-2
Could you please go over an explanation for Q2 using Euler's method?
For some reason I did this in my head by using the natural log and stuff, but I did manage to get -3 and 1.
I then looked at it again and knew I didn’t need that at all. At least I did it in my head :P
For the case 2 I got it by doing *x to both sides to get x^(x+3+1)=x^1 which same bases so it's x+4=1 which with basic algebra it goes to x=-3
That way was close to proving case 3 anything to an even power is positive. So if x^(x + 3) = 1 was figured out in cases 1 as +1 to any power odd or even (since +1 is an odd integer then 1 + 3 = 4 is an even integer) ... and 2 of odd integers -3 and 1 raised to an even power themselves, respectively. We are still missing one other odd integer raised to an even power from the negative integers after the x + 3 = 0 solution giving x = -3 of (-1)^(2) one of case 3! ...
Then you were starting to show case 3 proof by multiplying both equation sides by x to get x^(x + 3) times x = 1 times x ... Or ... x^(x + 4) = x that would have shown his case 3 result (-1)^(-1 + 4) = (-1) parallel to saying any integer positive or negative to an even integer power is a positive integer result! Good try! 👍
That's the bprp I wanna see, he gave us summer homework. 🤣🤣
😆
An other solution using log function we use the absolut value to have the function defind in R
(X+3) log(|X|) = 0 => X+3 = 0 => x = -3 or X = 1 or X = -1 (because X+3 is even)
That’s how I was thinking but forgot about the absolute value so just got positive 1. Looks right but the reasoning behind x=1 and x=-1 as solutions is because it makes ln|x| equal zero not because x+3 is even. Well I guess it’s true both ways but it definitely makes the natural log function zero.
Can you do the Spanish math college entrance exam of 2024, it’s rising a lot of controversy because of the difficulty
El de Madrid? A mí me han reventado con la EvAU de Mates jajajjaja
He is a great teacher
Just a quick question to the wonderful admin or anyone who reads this.
If I start learning maths from grade 10(other than previous grades), will I know enough to attempt engineering entrance exams?
Like the examples above and so on?
Thanks
Hellooooo. Hadnt u uploaded a video of P(x) ^ Q(x) = 1 back in the past? Where solutions where 2 3 4 5 6 7 and wolfram alpha was missing 2 of them? Pls, if i am right, could you link it? I cant find it :(
solution to Q1:
case 1:
1 to the power of any number is 1
so x+2 = 1
x = -2+1 = -1
case 2:
any number to the zeroth power is 1, so x+3=0
therefore, x = -3
case 3:
(-1)^even is 1, so lets put x+2=-1
so, x = -1-2 = -3
wait...no nevermind let's keep going
the exponent will be -3+3=0
so, (-1)^0 = 1? TRUE
since the solution to case 2 and 3 are the same, I will only write it once
so finally, x = (-1,-3)
solution to Q2:
(-1)^odd = -1, so we'll let x+2 = -1 => x = -1-2 = -3
putting that into the original, (-1)^(-3+3) = (-1)^0 = -1? FALSE
so no real solution
solution to Q3:
0 to any positive real number is 0
so, assuming x+2 = 0, x=-2
putting that into the original, 0^(-2+3) = 0^1 = 0? TRUE
so, x=-2
I just found a more straightaway solution to Q1 using logarithms
take ln on both sides and bring the power to the front to get (x+3)*ln(x+2) = 0
x+3 = 0
x = -3
ln(x+2) = 0
x+2 = 1
x = -2+1 = -1
so again, x=(-1,-3)
now I found a flaw in this
if you do this, you are limiting yourself from ever getting the answer from case 3
however, I found a solution and it is that you can let the argument be -1 and if x+3 is even, you can bring the power up to the exponent of -1 to get 1 inside the ln and finally get ln(1) = 0
we are just lucky to have case 2 and 3 yielding the same answer, otherwise we'd have another solution that we would've missed
Nice work man
@@Brid727thanks for the easy to understand and great solution, cannot thank you enough
@@Brid727 ..ln(x+2)=0 ... ln(anything)=0 will be undefined,
ur asking what exponent of '0' makes 'x+3' (since u set it equal to zero), since there exist no real solution its undefined 0^anything=0, 0^0 is undetermined/undefined
Basicly e^zero=0 (wich has no real solution)
however..
ln(x+3)=ln(x)ln(3), ln(3) is close to ln(e)=1[e^1=e] , 1.09... -> 1.09 * X≈0, for practical reasons we can assume the limit x as zero, but as a logarithm ln(zero) is undefined
..in case x would be infinitassible small but not zero...
@@Patrik6920 ln(1) = 0 laughing in the corner 😂
also please read what I wrote above and plug x=-1 in and if you have the slightest bit of knowledge in logarithms you'd know that ln(1)=0
I think you are confusing what I wrote with ln(0) which is clearly NOT what you get for plugging x=-1
Excellent
Nice video ❤
3 condition to get 1.
Any no power 0 is 1
1 power any no. Is 1
-1 power even no. Is 1
So. X= -3; 1;-1 are solns. In sequence.
Why should the exponent in case 1 be finite?
"anything to the power of zero, equals one".
Thus 1 = x^(x+3) is possible only when
x+3 = 0/That is x = -3.
I have a question. Is there anyway to solve this without logic? I imagine you could use the Lambert W function in some way, but I don't know how. If any commenter knows, feel free to educate me.
You could do e^((x+3)lnx) = e^0. Then (x+3)lnx = 0 x = -3 and x = 1, but you don't get the third answer.
@@acetylsalicylicacid You do get the third answer, actually. Log is only defined for non-negative arguments, now since x^(x+3) is 1, we know that it's positive, but we do not know whether or not x itself is positive or negative. So when you pull the exponent out front you need to include absolute value signs, so we actually get (x+3)ln|x| = 0, and ln|x|=0 gives us both 1 and -1.
@@phiefer3 Thank you so much! It's strange he didn't go over this method in the video.
does x=0 really work if we expand the equation into x^x * x^3 =1 ? Cuz I thought 0^0 is 1 (using limits), and 1*0=0?
Other example
-1 for 1^x rule
-3 for x^0 rule and (-1)^2n rule
1: x= -1 or -3
2: couldn’t figure this out lol
3: x= -2
Is it possible to solve using a rule ? Like its easy here since we have ..... = 1
Probably have to break out the productlog function to prove that one
Just use logs
Oh, i solved it by multiplying both sides by X to have X+4=1 and then X, then X=-3
My answers of the last 3 questions:
Q1: x = -1, -3
Q2: No solutions
Q3: x = -2
I dit it with logarithms:
X^(X + 3) = 1
(X+3)log(X) = 0
Either X = -3 or X = 1
For the Assignment
X = -3 only
(X+2)^x+3 = 1
(X+2)^x+3 = (x+2)⁰
When the base cancels
X+3 = 0
Then u make x subject
X = -3
X=-1 also works since it makes the base 1
2nd assignment requires (-1)^odd but for base to be -1 x would be -3 which would make exponent even and therefore not -1.
@@fifiwoof1969 -1 is not a solution for x in the assignment... U can take ur time and substitute x as -1 in it and see it won't give u 1 as result
@@engrkingsley6854 for the first assignment it DOES work. Take YOUR time.
@@fifiwoof1969 he already solve Those ones
@@engrkingsley6854 not the questions at the end he didn't - "assignment"
Putting log base x in both sides then simply solving it
Just trying to solving it from fundamentals
0^0 is 1. No one will judge you
You won't get the L'Hopitals limit in base "x" to the "x + 3" when x = 0 or x + 3 = 0. In those cases the (0)^(3) is 0. The (-3)^(0) will equal "1" he shows as case #2. ....
The one your brain was saying as a third solution was instead saying his case #3. We forget any integer base to an even power is a positive number result.
But the base is > 0 because the power is a real number, not natural anymore.
Q1: x=-1, x=-3
Q2: no real root
Q3: x=-2
Jinkies! He gave us homework! 😊
why the graphs y=x^(x+3) & y=1 has only one intersect x=1?
"The graph is dotted" when the base is negative. Plugging in x=-0.5 for example will give the root of a negative number.
-3
Can you teach me how can I sove this problem, please?
sqr(a)+sqr(ab)+sqr(abc)=12
sqr(b)+sqr(bc)+sqr(abc)=21
sqr(c)+sqr(ac)+sqr(abc)=30
Find: (a^2 + b^2 + c^2)
Q1: x= -3,x= -1
Q2: no solution
Q3: x= -2
First Look my guess is x = 1, -1 and -3
so simple🥱🥱🥱
I would just plug one (Obvious solution) and call it a day lol
wrong! Kindly note the function is only defined for x >0.
thr only solution is x = 1.
X=-3?
Q1: x = -1 or x = -3
Q2: no real solution
Q3: x = -2
Quite weird...
(x+2)^(x+3) = 1 --> x = -3 or -1
(x+2)^(x+3) = 0 --> x = -2
(x+2)^(x+3) = -1 --> Oh, no. 4 disgusting complex numbers... Stop this, get some help.
Those
ln(-1) = i(pi + 2 k pi)
stuff has come back...
(Like
ln(a+bi)
= ln(sqrt(a^2+b^2) exp(i atan(b/a)))
= ln(sqrt(a^2+b^2)) + i atan(b/a)
and something else...)
1 because 1^something = 1
-1 because (-1)^2n = 1
And -3 because something^0=1
Your second line is wrong. You need grouping symbols: (-1)^(2n) = 1, for n an integer.
First
If you want to _avoid_ the debate, then you can't just come out and say that the base can't be 0, because then you get me here in the comments debating it. Say that _maybe_ the base can be 0, we'll come back to it if necessary; solve x+3=0, find that the base is _not_ 0, and now you don't have to deal with it. Everybody's happy, the only real solutions are definitely 1, −1, and −3.
-3