<s>[INST] What to do with exponent with different base in denominator? I‘m struggling right now with a equation. The equation is
$$\frac{2^{(5x)}}{7^{(x+2)}} = 10.$$
The solution should be $4.076$ but I don‘t know how to solve this equation.
I came to the conclusion that I can simplify the equation so it‘s $\frac{32^x}{7^{(x+2)}}$. My problem is that I don‘t know how I can get rid of the denominator.
Thank you for your help.
 [/INST] You simplified $2^{5x}$ to $\left(2^5\right)^x=32^x$.

You can also simplify $7^{x+2}$ to $7^x\cdot7^2=49\cdot7^x$, using the exponent rule $a^{b+c}=a^b\cdot a^c$.
Thus, your equation simplifies to
$$
\begin{align*}
\frac{32^x}{49\cdot7^x}&=10\\
\frac{1}{49}\cdot\frac{32^x}{7^x}&=10\\
\frac{32^x}{7^x}&=490\\
\left(\frac{32}{7}\right)^x&=490\\
\ln\left(\left(\frac{32}{7}\right)^x\right)&=\ln(490)\\
x\ln\left(\frac{32}{7}\right)&=\ln(490)\\
x&=\frac{\ln(490)}{\ln\left(\frac{32}{7}\right)}\\
x&\approx4.076
\end{align*}
$$
</s>