How do you rewrite ${log}_{3} 8 - 4 {log}_{3} (x^{2})$ $Log_3 8 - 4log_3(x^2)$ as a single log?

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Answer 1 · 2015-10-30T11:37:46

$- 8 {log}_{3} 8 x$ $-8log_(3)8x$

By using laws of logs and exponents we get:

${log}_{3} 8 - 4 {log}_{3} x^{2} = {log}_{3} 8 - 8 {log}_{3} x$ $log_(3)8-4log_3x^2=log_(3)8-8log_3x$

$= {log}_{3} 8 - {log}_{3} x^{8}$ $=log_(3)8-log_3x^8$

$= {log}_{3} (\frac{8}{x^{8}})$ $=log_3(8/x^8)$

$= {log}_{3} 8 x^{- 8}$ $=log_(3)8x^(-8)$

$= - 8 {log}_{3} 8 x$ $=-8log_(3)8x$

(Note: There might be other valid ways to express it as well)

How do you rewrite log38−4log3(x2)Log_3 8 - 4log_3(x^2) as a single log?