If $\log2=m$ and $ \log3=n$, What is the answer of
$\large3^{2^x}=\large2^{3^x}$? $1)\log
mn\quad\quad\quad\quad\quad\quad2)\frac{m-n}{\log m-\log
n}\quad\quad\quad\quad\quad\quad3)\frac{\log \frac
mn}{m-n}\quad\quad\quad\quad\quad\quad4)\frac{m+n}{\log mn}$
Here is my approach:
To solve the equation I took logarithm in base $3$ of both sides of the equation:
$$2^x=\log_32^{3^x}=3^x\log_32=3^x\times\frac{m}{n}$$
So we have $(\frac23)^x=\frac mn$. by taking logarithm in base $\frac23$ we have:
$$x=\log_{\large\frac23}(\frac mn)=\frac{\log(\frac mn)}{\log(\frac23)}=\frac{\log \frac mn}{m-n}$$
My questin: Is there other approach to solve this problem?
Best Answer
You could take natural log. It will reduce the change of base steps
$\Rightarrow 2^x\log 3 = 3^x\log 2 \\\Rightarrow (\frac23)^x =\frac mn \\\Rightarrow x\log(\frac23) = x(m -n) = \log(\frac mn)\\ \Rightarrow \boxed{x = \frac{\log \frac mn}{m-n}} $