Question

log(base 16) x + log(base 4) x + log(base 2) x = 7.
Solve for X

Answer 1

\[\log_{16} x + \log_{4} x + \log_{2} x = 7\]

Answer 2

This is how I solved it, I believe it is correct, but I'm sure there is an easier way.
First convert to same base
\[\log _{16}x + \log_{4}x+\log_{2}x = 7\]\[\log_{2}x^{\frac{1}{4}}+\log_{2}x^{\frac{1}{2}} + \log_{2}x = 7\]
The combine into one log
\[\log_{2}(x^{\frac{1}{4}}+x^{\frac{1}{2}} + x ^1)= 7 => \log_{2}(x^{\frac{7}{4}})= 7\]

This means
\[2^7 = x^{\frac{7}{4}}\]

Raise both sides by 4/7 (to get x^1)

\[2^{7{\frac{4}{7}}} = x^1=> 2^4 = x => x = 16\]

Answer 3

oops, there is a typo. When combining the log the x's should be multiplied, not added.