How do you solve log_2 x + log_4 x + log_8 x + log_16 x = 25?

Mar 4, 2016

Solution $\implies x = {2}^{12}$

Explanation:

${\log}_{2} x + {\log}_{4} x + {\log}_{8} x + {\log}_{16} x = 25$
$\implies \frac{1}{\log} _ x 2 + \frac{1}{\log} _ x {2}^{2} + \frac{1}{\log} _ x {2}^{3} + \frac{1}{\log} _ x {2}^{4} = 25$
$\implies \frac{1}{\log} _ x 2 + \frac{1}{2 {\log}_{x} 2} + \frac{1}{3 {\log}_{x} 2} + \frac{1}{4 {\log}_{x} 2} = 25$
$\implies \frac{1}{\log} _ x 2 \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) = 25$
$\implies {\log}_{2} x \frac{12 + 6 + 4 + 3}{12} = 25$
$\implies {\log}_{2} x = 25 \cdot \frac{12}{25} = 12$
$\implies x = {2}^{12}$

Mar 4, 2016

We can start by converting all of the $\log$ terms to the same base using the rule;

${\log}_{c} b = \log \frac{b}{\log} c$

$\log \frac{x}{\log} 2 + \log \frac{x}{\log} 4 + \log \frac{x}{\log} 8 + \log \frac{x}{\log} 16 = 25$

Now we can pull $\log x$ out of the numerators.

$\log x \left(\frac{1}{\log} 2 + \frac{1}{\log} 4 + \frac{1}{\log} 8 + \frac{1}{\log} 16\right) = 25$

Notice that all of the denominators are powers of 2, so we can rewrite as;

$\log x \left(\frac{1}{\log} 2 + \frac{1}{\log} \left({2}^{2}\right) + \frac{1}{\log} \left({2}^{3}\right) + \frac{1}{\log} \left({2}^{4}\right)\right) = 25$

Using the identity, $\log \left({x}^{y}\right) = y \cdot \log x$, we can rewrite the denominators as multiples of $\log 2$.

$\log x \left(\frac{1}{\log} 2 + \frac{1}{2 \log 2} + \frac{1}{3 \log 2} + \frac{1}{4 \log 2}\right) = 25$

Now we can pull $\log 2$ out of the denominators.

$\log \frac{x}{\log} 2 \left(1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right) = 25$

We can rewrite all of the fractions in term of the lowest denominator.

$\log \frac{x}{\log} 2 \left(\frac{12}{12} + \frac{6}{12} + \frac{4}{12} + \frac{3}{12}\right) = 25$

Sum the terms in parenthesis.

$\log \frac{x}{\log} 2 \left(\frac{25}{12}\right) = 25$

We can combine all of the constant terms by multiplying both sides by the inverse of the fraction on the left hand side.

$\log \frac{x}{\log} 2 \left(\frac{25}{12}\right) \left(\frac{12}{25}\right) = 25 \left(\frac{12}{25}\right)$

Combine the $\log$ terms by reversing the process we did in step one and eliminate all of the constant terms that cancel.

${\log}_{2} x = 12$

Using the definition of a logarithm, we write this statement in terms of $x$.

$x = {2}^{12}$

Solve.

$x = 4096$