# How do you solve this logarithm question.

#### 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????

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#### Sudharaka

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,

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

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#### e^(i*pi)

MHF Hall of Honor
Completely wrong here, see above and below posts.

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

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#### Sudharaka

$$\displaystyle {\color{red}\log_2(a) \cdot \ln(x) + \log_2(x) \cdot \ln(a) = \ln(2) + 2\ln(a)}$$

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

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

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

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

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

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; $$\displaystyle log(A+B)\neq{logA+logB}$$

• e^(i*pi)

#### e^(i*pi)

MHF Hall of Honor
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 e^(i*pi),

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

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

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

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

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

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

$$\displaystyle \Rightarrow x = 4$$

• arijit2005

#### Sudharaka

Hi everyone,

Another approch:

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

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

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

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

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

$$\displaystyle \Rightarrow{\log_2x=2}$$

$$\displaystyle \Rightarrow{x=4}$$

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• Rapha and arijit2005

#### arijit2005

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

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

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

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

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

$$\displaystyle \Rightarrow x = 4$$

WOW..(Surprised) Thanks.. That's a nice way..(Happy) Why the hell didn't I think of that???(Headbang) I need more practice.(Thinking)

#### arijit2005

Hi everyone,

Another approch:

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

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

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

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

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

$$\displaystyle \Rightarrow{\log_2x=2}$$

$$\displaystyle \Rightarrow{x=4}$$

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

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

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

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

MHF Hall of Honor
Hello arijit2005
Hey... Thanks a lot man!!! No.. I haven't been introduced to this new formula $$\displaystyle \huge{x^{log_{a}y}=y^{log_{b}x}}$$

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

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

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

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

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

$$\displaystyle = (a^b)^c$$

$$\displaystyle = y^c$$

$$\displaystyle =y^{\log_ax}$$