# Thread: How do you solve this logarithm question.

1. ## How do you solve this logarithm question.

FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -

x^[log a to the base 2] + a^[log x to the base 2] = 2a^2

I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????

2. Originally Posted by arijit2005
FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -

x^[log a to the base 2] + a^[log x to the base 2] = 2a^2

I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
Dear arijit2005,

Although you may not be familiar there's a logarithm identity,

$\huge{x^{log_{a}y}=y^{log_{a}x}}$

3. Completely wrong here, see above and below posts.

It might be prudent for this post to be deleted since it contributes nothing

4. Originally Posted by e^(i*pi)
${\color{red}\log_2(a) \cdot \ln(x) + \log_2(x) \cdot \ln(a) = \ln(2) + 2\ln(a)}$

By the change of base rule: $\log_2(x) = \frac{\ln(x)}{\ln(2)}$

$\frac{\ln(a)}{\ln(2)} \cdot \ln(x) + \frac{\ln(x)}{\ln(2)} \cdot \ln(a) = \ln(2)+2\ln(a)$

$\ln(a)\ln(x) + \ln(x)\ln(a) = \ln(2)(\ln(2)+2\ln(a))$

$2\ln(a)\ln(x) = (\ln(2))^2 + 2\ln(a)\ln(2)$

$\ln(x) = \frac{(\ln(2))^2 + 2\ln(a)\ln(2)}{2\ln(a)}$

Should be easy enough to find x in terms of a from there. I have used base e but any base will work
Dear e^(i*pi),

There's a mistake in your first step; $log(A+B)\neq{logA+logB}$

5. Originally Posted by arijit2005
FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -

x^[log a to the base 2] + a^[log x to the base 2] = 2a^2

I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
Originally Posted by Sudharaka
Dear e^(i*pi),

There's a mistake in your first step; $log(A+B)\neq{logA+logB}$
Of course you're right, how on Earth did I not spot that >.<

6. Hello arijit2005
Originally Posted by arijit2005
FIND OUT THE VALUE OF "X" FROM THE EQUATION GIVEN BELOW: -

x^[log a to the base 2] + a^[log x to the base 2] = 2a^2

I tried taking Log to the bast 10 to both sides, but it didn't work out. Any idea how to do it????
We have to solve for $x$:
$x^{\log_2a}+a^{\log_2x}=2a^2$
If in doubt, get rid of logs as soon as you can - they're nasty things! So let $b = \log_2a$. Then $a=2^b$. So the equation becomes:
$x^b + \big(2^b\big)^{\log_2x} = 2\big(2^b\big)^2$

$\Rightarrow x^b + 2^{b\log_2x} = 2\cdot2^{2b}$

$\Rightarrow x^b + 2^{\log_2(x^b)} = 2\cdot2^{2b}$

$\Rightarrow x^b + x^b = 2\cdot(2^2)^b$

$\Rightarrow 2x^b = 2\cdot4^b$

$\Rightarrow x = 4$

7. Hi everyone,

Another approch:

If you use the logarithmic identity that I had mentioned in my previous post; $\huge{x^{log_{a}y}=y^{log_{a}x}}$

$x^{\log_2a}+a^{\log_2x}=2a^2$

$\Rightarrow{a^{\log_2x}+a^{\log_2x}=2a^2}$

$\Rightarrow{2a^{\log_2x}=2a^2}$

$\Rightarrow{a^{\log_2x}=a^2}$

$\Rightarrow{\log_2x=2}$

$\Rightarrow{x=4}$

Hello arijit2005We have to solve for $x$:
$x^{\log_2a}+a^{\log_2x}=2a^2$
If in doubt, get rid of logs as soon as you can - they're nasty things! So let $b = \log_2a$. Then $a=2^b$. So the equation becomes:
$x^b + \big(2^b\big)^{\log_2x} = 2\big(2^b\big)^2$

$\Rightarrow x^b + 2^{b\log_2x} = 2\cdot2^{2b}$

$\Rightarrow x^b + 2^{\log_2(x^b)} = 2\cdot2^{2b}$

$\Rightarrow x^b + x^b = 2\cdot(2^2)^b$

$\Rightarrow 2x^b = 2\cdot4^b$

$\Rightarrow x = 4$

WOW.. Thanks.. That's a nice way.. Why the hell didn't I think of that??? I need more practice.

9. Originally Posted by Sudharaka
Hi everyone,

Another approch:

If you use the logarithmic identity that I had mentioned in my previous post; $\huge{x^{log_{a}y}=y^{log_{b}x}}$

$x^{\log_2a}+a^{\log_2x}=2a^2$

$\Rightarrow{a^{\log_2x}+a^{\log_2x}=2a^2}$

$\Rightarrow{2a^{\log_2x}=2a^2}$

$\Rightarrow{a^{\log_2x}=a^2}$

$\Rightarrow{\log_2x=2}$

$\Rightarrow{x=4}$

Hey... Thanks a lot man!!! No.. I haven't been introduced to this new formula $\huge{x^{log_{a}y}=y^{log_{b}x}}$

By the way, is it $\huge{x^{log_{a}y}=y^{log_{b}x}}$ or $\huge{x^{log_{a}y}=y^{log_{a}x}}$???

Will base "a" change?? You actually wrote $\huge{y^{log_{b}x}}$

Are you sure the $a$ becomes $b$???

10. Hello arijit2005
Originally Posted by arijit2005
Hey... Thanks a lot man!!! No.. I haven't been introduced to this new formula $\huge{x^{log_{a}y}=y^{log_{b}x}}$

By the way, is it $\huge{x^{log_{a}y}=y^{log_{b}x}}$ or $\huge{x^{log_{a}y}=y^{log_{a}x}}$???

Will base "a" change?? You actually wrote $\huge{y^{log_{b}x}}$

Are you sure the $a$ becomes $b$???
Yes, it should be
$x^{\log_ay}=y^{\log_ax}$
You can prove this using the same substitution that I used (so that you can get rid of those nasty logs!):
Let $b = \log_ay$ and $c = \log_ax$

Then $a^b = y$ and $a^c = x$

So $x^{\log_ay} = x^b$
$=(a^c)^b$

$= (a^b)^c$

$= y^c$

$=y^{\log_ax}$