Question

# Condense them to the same base before solving for x log16 (x) +log4 (x) +log2 (x) =7

Logarithms
Condense them to the same base before solving for x $$\displaystyle{\log{{16}}}{\left({x}\right)}+{\log{{4}}}{\left({x}\right)}+{\log{{2}}}{\left({x}\right)}={7}$$

$$\displaystyle{\log{{16}}}{\left({x}\right)}={\left(\frac{{1}}{{4}}\right)}{\log{{2}}}{\left({x}\right)}$$
$$\displaystyle{\log{{4}}}{\left({x}\right)}={\left(\frac{{1}}{{2}}\right)}{\log{{2}}}{\left({x}\right)}$$
so you hav $$\displaystyle{\left({1}+\frac{{1}}{{2}}+\frac{{1}}{{4}}\right)}{\log{{2}}}{\left({x}\right)}={7}$$
$$\displaystyle{1.75}\cdot{\log{{2}}}{\left({x}\right)}={7}$$
$$\displaystyle{\log{{2}}}{\left({x}\right)}=\frac{{7}}{{1.75}}={4}$$
$$\displaystyle{x}={2}^{{4}}={16}$$