common logarithms: solve 2^(log2x) = 3^(log3x)

[padamarthur]

2^log2x=3^log3x
log2^log2x=log3^log3x
(log2X)(log2)=(log3x)(log3)
(log2 + logX)(log2)=(log3 + logX)(log3)

Im not exactly sure where to go from here

 

[pka]

2^log2x=3^log3x
log2^log2x=log3^log3x
(log2X)(log2)=(log3x)(log3)
(log2 + logX)(log2)=(log3 + logX)(log3)

$\displaystyle (\log(2) + \log(X))(\log(2))=(\log(3) + \log(X)(\log(3))$

$\displaystyle (\log(X))(\log(2)-(\log(3))=(\log^2(3) -(\log^2(2)$

$\displaystyle \log(X)=\dfrac{\log^3(3) -\log^2(2)}{\log(2)-(\log(3)}$

 

[lookagain]

$\displaystyle (\log(2) + \log(X))(\log(2))=(\log(3) + \log(X)(\log(3))$

$\displaystyle (\log(X))(\log(2)-(\log(3))=(\log^2(3) -(\log^2(2)$

$\displaystyle \log(X)=\dfrac{\log^3(3) -\log^2(2)}{\log(2)-(\log(3)}$

There are a few typos above.


$\displaystyle (\log(2) + \log(X))(\log(2)) = (\log(3) + \log(X))(\log(3))$


$\displaystyle (\log(X))(\log(2) - (\log(3)) = (\log^2(3) - \log^2(2))$


$\displaystyle \log(X)=\dfrac{\log^2(3) - \log^2(2)}{\log(2) - \log(3)}$


$\displaystyle \log(X)=- \ \dfrac{\log^2(3) - \log^2(2)}{\log(3) - \log(2)} \ \ \ \ \ \ \ \$ <---- And continuing...


$\displaystyle \log(X)=- \ \dfrac{(\log(3) + \log(2))(\log(3) - \log(2))}{\log(3) - \log(2)}$


And continue...

 

[padamarthur]

$\displaystyle (\log(2) + \log(X))(\log(2)) = (\log(3) + \log(X))(\log(3))$

Suggestion to simplify plus speed up:

a = log(2), b = log(3), c = log(X) ; above becomes:

a(a + c) = b(b + c)

a^2 + ac = b^2 + bc

ac - bc = -a^2 + b^2

c(a - b) = (-a + b)(a + b)

c(a - b) = -(a - b)(a + b)

c = -(a + b)

Wrap up by substituting back in...

Save time on timed tests, plus on pencil/eraser wear:wink:

so to finish; logX =-(log2 + log3)
logX =-(log6)
logX = log6-1
therfore X = [1][/6]

 

[lookagain]

so to finish; logX =-(log2 + log3)

logX =-(log6)

logX = log6-1 $\displaystyle \ \ \ \ \ \$ The "-1" has to be shown as a superscript, or write it as logX = log(6^(-1)).

therfore X = [1][/6] $\displaystyle \ \ \ \ \ \$ Drop the brackets. X = 1/6.

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