Question

By using the laws of logarithm, find the value of log(81/8)-2log(3/2)+3log(2/3)+log(3/4)​

Answer 1

Answer:

$\huge\boxed{ \green{ 2 log( \frac{9}{2} ) }}$

Soln refers to the attachment

Formula used here :

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$\boxed{ \red{\bold{ log(x) + log(y) = log(xy) }}}$

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$\boxed{ \purple{\bold{ log(x) - log(y) = log( \frac{x}{y} ) }}}$

◆

$\boxed{ \bold{ \blue{a \: log(b) = log( {b}^{a} ) }}}$

Answer 2

$\pink{\bold{\underline{\underline{Question:-}}}}$

find the value of log(81/8)-2log(3/2)+3log(2/3)+log(3/4) ?

$\green{\bold{\underline{\underline{Step\:by\:step\:explanation:}}}}$

$\red{\bold{we\:know\:that:--}}$

● $\orange{\bold{Product\:Rule\:Law:}}$

loga (MN) = loga M + loga N

● $\pink{\bold{Quotient\:Rule\:Law:}}$

loga (M/N) = loga M - loga N

● $\green{\bold{Power\:Rule\:Law:}}$

IogaM^(n) = n Ioga M

Using all in Question now ,

log(81/8)-2log(3/2)+3log(2/3)+log(3/4)

$log( \frac{81}{8} ) - log( \frac{3}{2} ) ^{2} + log( \frac{2}{3} )^{3} + log( \frac{3}{4} ) \\ \\ \implies \: log( \frac{81}{8} ) - log( \frac{9}{4} \times \frac{8}{27} \times \frac{3}{4} ) \\ \\ \implies \: log( \frac{81}{8} ) - log( \frac{1}{2} ) \\ \\ \implies \: log( \frac{ \frac{81}{8} }{ \frac{1}{2} } ) \\ \\ \implies \: log( \frac{81}{8} \times \frac{2}{1} ) \\ \\ \implies \: log( \frac{81}{4} ) \\ \\ \implies \: log( \frac{9}{2} )^{2}$

$\textsf{</strong><strong>our</strong><strong> </strong><strong>answer</strong><strong> </strong><strong>is</strong><strong>:</strong><strong>-</strong><strong>}$

$\bold{\boxed{\huge{\boxed{\orange{\small{\boxed{\huge{\red{\bold{\:</strong><strong>2 log( \frac{9}{2} )</strong><strong>}}}}}}}}}}$

$\huge\underline\mathfrak\green{Hope\:it\:Helps\:You}$