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avatar+261 

\(\log{x}=\frac{\log{2}\log{3}}{\log{4}}\)

 

Solve for x

 

The numerator is giving me issues as it has two logs in it. Can't seem to simplify it after raising both sides to 10, any tips?

Quazars  Jan 13, 2018
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4+0 Answers

 #1
avatar+261 
+1

(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

Quazars  Jan 13, 2018
 #2
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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)

 

Guest Jan 13, 2018
 #3
avatar+261 
+1

Can you do that even if the log is in base 10? As in do you not have to raise it to the power of 10 instead?

 

How do you go from step 2 to 3? As in how do you simplify e^((log(2) log(3))/(log(4))) to sqrt(3)?

Quazars  Jan 13, 2018
 #4
avatar+26677 
+2

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

 

Also, (1/2) log 3 is log 31/2

 

Hence x = 3 1/2

Alan  Jan 13, 2018

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