Question

Guys I'm stuck! can someone help me with this log function?

Log X^2 + Log X^6
---------------------------- = 3
Log 6X

Answer 1

https://www.brainfuse.com/jsp/alc/resource.jsp?s=gre&c=36985&cc=108829 i dont know if this helps i just googled your answer and found this. ;)

Answer 2

\[\begin{cases}
\log x^n=n\log x\\
\log(mn)=\log m+\log n\end{cases}\implies\frac{\log x^2+\log x^6}{\log 6x}=\frac{2\log x+6\log x}{\log6+\log x}\]so you can rewrite the original equation as
\[\frac{8\log x}{\log 6+\log x}=3\implies 8\log x=3\log6+3\log x\implies 5\log x=3\log 6\]Can you find \(x\) from here?

Answer 3

When do we expand the log using the log laws and when to we compress it? I'm so confused as to me we should either expand the equation or compress it...

Answer 4

Expanding and compressing at the same time makes no sense to me

Answer 5

(log x^2 + log x^6)/(log 6x) = 3
=> log (x^2)*(x^6) = 3 log 6x
=> log (x^8) = log (6x)^3
=> log (x^8) = log (216x^3)

x^8 = 216x^3
x^5 = 216
x = (log 216)/(log 5)
x = (3 log 6)/(log 5)

Answer 6

Please get back to me it's urgent