Question

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

Logarithms
ANSWERED
asked 2020-10-20
Condense them to the same base before solving for x \(\displaystyle{\log{{16}}}{\left({x}\right)}+{\log{{4}}}{\left({x}\right)}+{\log{{2}}}{\left({x}\right)}={7}\)

Answers (1)

2020-10-21
\(\displaystyle{\log{{16}}}{\left({x}\right)}={\left(\frac{{1}}{{4}}\right)}{\log{{2}}}{\left({x}\right)}\)
\(\displaystyle{\log{{4}}}{\left({x}\right)}={\left(\frac{{1}}{{2}}\right)}{\log{{2}}}{\left({x}\right)}\)
so you hav \(\displaystyle{\left({1}+\frac{{1}}{{2}}+\frac{{1}}{{4}}\right)}{\log{{2}}}{\left({x}\right)}={7}\)
\(\displaystyle{1.75}\cdot{\log{{2}}}{\left({x}\right)}={7}\)
\(\displaystyle{\log{{2}}}{\left({x}\right)}=\frac{{7}}{{1.75}}={4}\)
\(\displaystyle{x}={2}^{{4}}={16}\)
0
 
Best answer

expert advice

Need a better answer?
...