How do you condense ${log}_{4} {(x y)}^{3} - {log}_{4} (x y)$ $log_4(xy)^3-log_4(xy)$ ?

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Answer 1 · 2018-07-06T15:22:30

$\Rightarrow {log (x y)}^{2}$ $color(cyan )(=> log (xy)^2$

${log}_{4} {(x y)}^{3} - {log}_{4} (x y)$ $log_4 (xy)^3 - log_4 (xy)$

${log x}^{n} = b log x$ $log x^n = blog x$

$\Rightarrow 3 {log}_{4} (x y) - {log}_{4} (x y)$ $=> 3 log_4 (xy) - log_4 (xy)$

$\Rightarrow 2 log (x y)$ $=> 2 log (xy)$

$\Rightarrow {log (x y)}^{2}$ $=> log (xy)^2$

Answer 2 · 2018-07-06T15:29:45

$2 log (x y) = {log (x y)}^{2}$ $2log(xy)=log(xy)^2$

Given: ${log}_{4} {(x y)}^{3} - {log}_{4} (x y)$ $log_4(xy)^3-log_4(xy)$

Use the fact that ${log}_{b} m^{n} = n {log}_{b} m$ $log_bm^n=nlog_bm$ .

$= 3 {log}_{4} (x y) - {log}_{4} (x y)$ $=3log_4(xy)-log_4(xy)$

Let $u = {log}_{4} (x y)$ $u=log_4(xy)$ .

$= 3 u - u$ $=3u-u$

$= 2 u$ $=2u$

Substitute back $u = {log}_{4} (x y)$ $u=log_4(xy)$ to get:

$= 2 log (x y)$ $=2log(xy)$

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

$= {log (x y)}^{2}$ $=log(xy)^2$

How do you condense log4(xy)3−log4(xy)log_4(xy)^3-log_4(xy)?