TITLE: Find $x$ in $4^x+6^x=9^x$
QUESTION [5 upvotes]: Divide The whole equation by $4^x$
$$1+\left (\frac 32\right)^x=\left(\frac 32\right)^{2x}$$
Solving the quadratic equation gives $$\left (\frac 32\right)^x=\frac{1+\sqrt 5}{2}$$
NOTE: the other value was not considered since exponential is always positive
This gives us the value of $x$ as 
$$x=\frac{\ln(1+\sqrt 5)-\ln 2}{\ln 3-\ln 2}$$
But the answer given is $\frac{\ln(\sqrt 5-1)-\ln 2}{\ln 2-\ln 3}$
What is going wrong?

REPLY [13 votes]: Note that
$${1+\sqrt 5}=(1+\sqrt 5)\frac{\sqrt 5 -1}{\sqrt 5 -1}=\frac4{\sqrt 5 -1}$$
therefore the two expressions are equivalent, indeed
$$\frac{\ln(1+\sqrt 5)-\ln 2}{\ln 3-\ln 2}=\frac{\ln\left(\frac4{\sqrt 5 -1}\right)-\ln 2}{\ln 3-\ln 2}=\frac{\ln4-\ln(\sqrt 5 -1)-\ln 2}{\ln 3-\ln 2}=$$
$$=\frac{2\ln2-\ln(\sqrt 5 -1)-\ln 2}{\ln 3-\ln 2}=\frac{-\ln(\sqrt 5 -1)+\ln 2}{\ln 3-\ln 2}=\frac{\ln(\sqrt 5-1)-\ln 2}{\ln 2-\ln 3}$$