Question

Please
$7 log( \frac{15}{16} ) + 6 log \frac{8}{3} + 5 log( \frac{2}{5} ) + log( \frac{32}{25} )$
solve this problem ​

Answer 1

$\huge \underline \bold \green{QUESTION}:$

$7 log( \frac{15}{16} ) + 6 log \frac({8}{3} )+ 5 log( \frac{2}{5} ) + log( \frac{32}{25} )$

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$\huge\underline\bold \red{final \: answer}$

${ \bold{the \: simplified \: form \: is \: log(3)} }$

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$\boxed{\underline\bold \red{formulas \: used}}$

$:⇉m log( \alpha ) = log( { \alpha }^{m} ) \\:⇉ log( \alpha ) + log( \beta ) + </p><p>log( \gamma ) = log( \alpha \beta \gamma )$

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$\huge \underline \bold \red{SOLUTION}:$

$= > 7 log( \frac{15}{16} ) + 6 log \frac({8}{3}) + 5 log( \frac{2}{5} ) + log( \frac{32}{25} ) \\ = > log { (\frac{15}{16}) }^{7} + log{( \frac{8}{3}) }^{6} + log {( \frac{2}{5}) }^{5} + log { \frac{32}{25} } \\ = > log {\huge{(}} \frac{ {15}^{7} }{ {16}^{7} } \times \frac{ {8}^{6} }{ {3}^{6} } \times \frac{ {2}^{5}}{ {5}^{5} } \times \frac{32}{25} \huge{)} \\ = > log{ \huge{(}} \frac{ {(3 \times 5)}^{7} \times {8}^{6} \times {2}^{5} \times {2}^{5} }{ {(2 \times 8)}^{7} \times {3}^{6} \times {5}^{5} \times {5}^{2} } \huge{)} \\ \: = > log{ \huge{(}} \frac{3 \times {2}^{3} }{8} \huge{)} \\ \: \: \: \: \: \: \: \: \: \: \: { \boxed{ \green{= > log(3)}}}$

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