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Answer 1 · 2017-06-05T08:08:03

I got

$log [(\frac{16}{3}) \cdot (\frac{2}{9})]$ $log[(16/3)*(2/9)]$

We know the properties of log:

$log a + log b = log (a \cdot b)$ $log a + log b = log (a*b)$

$log a - log b = log (\frac{a}{b})$ $log a - log b = log(a/b)$

$a log b = {log b}^{a}$ $a log b = log b^a$

We have:

$(- log 3 + 2 log 4) - (- log 2 + 2 log 3)$ $(-log 3 + 2 log 4) - (- log 2+ 2 log 3)$

Rearranging

$(2 log 4 - log 3) - (2 log 3 - log 2)$ $(2 log 4 - log 3)-(2 log 3 -log 2)$

Applying properties, solving parentheses first

$({log 4}^{2} - log 3) - ({log 3}^{2} - log 2)$ $(log 4^2 - log 3)-(log 3^2 - log 2)$

$[log (\frac{4^{2}}{3})] - [log (\frac{3^{2}}{2})]$ $[log(4^2/3)]-[log(3^2/2)]$

$[log (\frac{16}{3})] - [log (\frac{9}{2})]$ $[log(16/3)]-[log(9/2)]$

Again it is of the form $log a - log b$ $log a- log b$

$log (\frac{\frac{16}{3}}{\frac{9}{2}})$ $log((16/3)/(9/2))$

Simplifying

$log [(\frac{16}{3}) \cdot (\frac{2}{9})]$ $log[(16/3)*(2/9)]$