How do you solve this logarithm question.

May 2010
18
1
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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Dec 2009
872
381
1111
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,

\(\displaystyle \huge{x^{log_{a}y}=y^{log_{a}x}}\)

This will help you to solve the problem.
 
Last edited:

e^(i*pi)

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Feb 2009
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Completely wrong here, see above and below posts.

It might be prudent for this post to be deleted since it contributes nothing
 
Last edited:
Dec 2009
872
381
1111
\(\displaystyle {\color{red}\log_2(a) \cdot \ln(x) + \log_2(x) \cdot \ln(a) = \ln(2) + 2\ln(a)}\)

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; \(\displaystyle log(A+B)\neq{logA+logB}\)
 
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e^(i*pi)

MHF Hall of Honor
Feb 2009
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West Midlands, England
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 >.<
 

Grandad

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Dec 2008
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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:
\(\displaystyle x^b + \big(2^b\big)^{\log_2x} = 2\big(2^b\big)^2\)

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

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

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

\(\displaystyle \Rightarrow 2x^b = 2\cdot4^b\)

\(\displaystyle \Rightarrow x = 4\)
Grandad
 
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Dec 2009
872
381
1111
Hi everyone,

Another approch:

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

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

\(\displaystyle \Rightarrow{a^{\log_2x}+a^{\log_2x}=2a^2}\)

\(\displaystyle \Rightarrow{2a^{\log_2x}=2a^2}\)

\(\displaystyle \Rightarrow{a^{\log_2x}=a^2}\)

\(\displaystyle \Rightarrow{\log_2x=2}\)

\(\displaystyle \Rightarrow{x=4}\)

Hope this will help you.
 
Last edited:
May 2010
18
1
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:
\(\displaystyle x^b + \big(2^b\big)^{\log_2x} = 2\big(2^b\big)^2\)

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

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

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

\(\displaystyle \Rightarrow 2x^b = 2\cdot4^b\)

\(\displaystyle \Rightarrow x = 4\)
Grandad

WOW..(Surprised) Thanks.. That's a nice way..(Happy) Why the hell didn't I think of that???(Headbang) I need more practice.(Thinking)
 
May 2010
18
1
Hi everyone,

Another approch:

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

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

\(\displaystyle \Rightarrow{a^{\log_2x}+a^{\log_2x}=2a^2}\)

\(\displaystyle \Rightarrow{2a^{\log_2x}=2a^2}\)

\(\displaystyle \Rightarrow{a^{\log_2x}=a^2}\)

\(\displaystyle \Rightarrow{\log_2x=2}\)

\(\displaystyle \Rightarrow{x=4}\)

Hope this will help you.

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 \(\displaystyle b\)???
 

Grandad

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Dec 2008
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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 \(\displaystyle x^{\log_ay} = x^b\)
\(\displaystyle =(a^c)^b\)

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

\(\displaystyle = y^c\)

\(\displaystyle =y^{\log_ax}\)
Grandad