How do you solve Log_2 x + log_4 x + log_8 x = 11 ?

Mar 8, 2018

$x = 64$

Explanation:

As ${\log}_{a} b = \log \frac{b}{\log} a$, we can write ${\log}_{2} x + {\log}_{4} x + {\log}_{8} x = 11$ as

$\log \frac{x}{\log} 2 + \log \frac{x}{\log} 4 + \log \frac{x}{\log} 8 = 11$

or $\log \frac{x}{\log} 2 + \log \frac{x}{2 \log 2} + \log \frac{x}{3 \log 2} = 11$

or $\log \frac{x}{\log} 2 \left(1 + \frac{1}{2} + \frac{1}{3}\right) = 11$

or $\log \frac{x}{\log} 2 \left(\frac{6 + 3 + 2}{6}\right) = 11$

or $\log \frac{x}{\log} 2 \left(\frac{11}{6}\right) = 11$

or $\log \frac{x}{\log} 2 = 11 \times \frac{6}{11} = 6$

i.e. $\log x = 6 \log 2 = \log {2}^{6}$

and $x = {2}^{6} = 64$