Logarithmic equation

(If someone could help me with the Latex images just a bit bigger that'd be great, not sure why but they come up a little tiny)

I managed to get the answer myself, posting it incase anyone is curious:

 

Start off by changing the log from base 10 to base 2 since 2 and 4 happen to give a integer power of 2

 

\(\frac{\log_{2}{x}}{\log_{2}{10}}=\left( \frac{\log_{2}{2}}{\log_{2}{10}}\frac{\log_{2}{3}}{\log_{2}{10}} \right)\frac{\log_{2}{4}}{\log_{2}{10}}\)

 

\(\frac{\log_{2}{x}}{\log_{2}{10}}=\left( \frac{1}{\log_{2}{10}}\frac{\log_{2}{3}}{\log_{2}{10}} \frac{\log_{2}{10}}{2}\right)\)

 

 

\(\log_{2}{x}=\frac{\log_{2}{3}}{2}\)

 

\(x=2^{\frac{1}{2}\log_{2}{3}}\)

 

\(x=\sqrt{2^{\log_{2}{3}}}\)

 

\(x=\sqrt{3}\)

 

 

Any other methods are very welcome

Solve for x:
log(x) = (log(2) log(3))/(log(4))

Cancel logarithms by taking exp of both sides:

x = e^((log(2) log(3))/(log(4)))

 

x =sqrt(3)

 

Note that log 4 is log 2², which is 2log 2, Hence the log 2 terms cancel.

 

Also, (1/2) log 3 is log 3^1/2

 

Hence x = 3 ^1/2