How do you write as a single log for log_460 - log_44 + log_4x ?

Dec 11, 2015

I think you intended ${\log}_{4} 60 - {\log}_{4} 4 + {\log}_{4} x$.

If so, then using the basic properties of logs this simplifies to:

${\log}_{4} 15 x$

Explanation:

In general, ${\log}_{a} b + {\log}_{a} c = {\log}_{a} b c$

and ${\log}_{a} b - {\log}_{a} c = {\log}_{a} \left(\frac{b}{c}\right)$

So:

${\log}_{4} 60 - {\log}_{4} 4 + {\log}_{4} x$

$= {\log}_{4} \left(\frac{60}{4}\right) + {\log}_{4} x$

$= {\log}_{4} 15 + {\log}_{4} x$

$= {\log}_{4} 15 x$