[Algebra]
Problem: Determine all values of $x$ for which $2^{\log _{10}\left(x^{2}\right)}=3\left(2^{1+\log _{10} x}\right)+16$.

Solution: Using logarithm rules, the given equation is equivalent to $2^{2 \log _{10} x}=3\left(2 \cdot 2^{\log _{10} x}\right)+16$ or $\left(2^{\log _{10} x}\right)^{2}=6 \cdot 2^{\log _{10} x}+16$.

Set $u=2^{\log _{10} x}$. Then the equation becomes $u^{2}=6 u+16$ or $u^{2}-6 u-16=0$.

Factoring, we obtain $(u-8)(u+2)=0$ and so $u=8$ or $u=-2$.

Since $2^{a}>0$ for any real number $a$, then $u>0$ and so we can reject the possibility that $u=-2$.

Thus, $u=2^{\log _{10} x}=8$ which means that $\log _{10} x=3$.

Therefore, $x=1000$.

Final answer: 1000