Logs!

$\large\log_{2}x +\log_{4}x = 3$ $\large\log_{4}x = \dfrac{1}{2} \log_{2}x$ $\large\log_{2}{x}+\dfrac{1}{2}\log_{2}x = 3$ $\large\dfrac{3}{2}\log_{2}x = 3$ $\large\log_{2}x = 2$ $\large x=4$

$\log _4 x + \log _2 x = 3$

$\log _4 x = 3 - \log _2 x$

$\Rightarrow x = 4^{3 - \log _2 x }$

$x = \frac{4^3} {4^{ \log _2 x }}$

$x = \frac{4^3} {2^{2 \cdot \log _2 x }}$

$x = \frac{4^3} {2^{ \log _2 x^2 }}$

$x = \frac{4^3} { x^2 }$

$x^3 = 4^3$

$x = 4$

$\boxed{ \Rightarrow \log _2 4 + \log _4 4 = 3 }$

$log_{2}x+log_{4}x=3$

If we change base 4 to base 2 (by using the change-of-base formula), we get

$log_{2}x+\frac{log_{2}x}{log_{2}4}=3$

Multiply both sides by two, and you then have

$log_{2}x=2 \Leftrightarrow x=4$

ln x = 3 *(ln 3 * ln 4) / (ln 3 + ln 4), x = 4

Rewrite as: log x / log 2 + log x / log 4 = 3

Multiply both sides by log 2 * log 4

log 4 * log x + log 2 * log x = 3 * log 2 * log 4

Factor

log x ( log 2 + log 4) = 3 * log 2 * log 4

Divide by the sum

log x = (3 * log 2 * log 4 ) / (log 2 + log 4)

By the rules of logs 3*log 2 = log 2^3 = log 8

AND log 2 + log 4 = log (2*4) = log 8

which leaves log 8 * log 4 / log 8 = log 4

log x = log 4 ==> x = 4

Logs, 4 word, 4 = answer :)