How do you condense $log_4(xy)^3-log_4(xy)$ ?

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Answer 1 · 2018-07-06T15:22:30

$color(cyan )(=> log (xy)^2$

$log_4 (xy)^3 - log_4 (xy)$

$log x^n = blog x$

$=> 3 log_4 (xy) - log_4 (xy)$

$=> 2 log (xy)$

$=> log (xy)^2$

Answer 2 · 2018-07-06T15:29:45

$2log(xy)=log(xy)^2$

Given: $log_4(xy)^3-log_4(xy)$

Use the fact that $log_bm^n=nlog_bm$ .

$=3log_4(xy)-log_4(xy)$

Let $u=log_4(xy)$ .

$=3u-u$

$=2u$

Substitute back $u=log_4(xy)$ to get:

$=2log(xy)$

Alternatively, we can bring the exponent back, and it will give us:

$=log(xy)^2$

How do you condense log_4(xy)^3-log_4(xy)log_4(xy)^3-log_4(xy)?