Question

Show that 2 log 5/8+ log 128/125+ log 5/2=0​

Answer 1

Question :

Show that : 2 log (5/8) + log (128/125) + log (5/2) = 0

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Explaination :

$\tt{ \implies \:LHS : 2 \: log \: \bigg( \dfrac{5}{8} \bigg) + log \: \bigg( \dfrac{128}{125} \bigg) + log \: \bigg ( \dfrac{5}{2} \bigg)}$

$\tt \implies \: log \: \bigg( \dfrac{5}{8} \bigg) ^{2} + log \: \bigg( \dfrac{128}{125} \bigg ) + log \: \bigg( \dfrac{5}{2} \bigg) \qquad \qquad \pmb{ \bigg \lgroup\because \: b.log \: a = log \: {(a)}^{b} } \bigg \rgroup$

$\tt \implies \: log \: \bigg (\dfrac{25}{64} \bigg) + log \: \: \bigg(\dfrac{5}{2} \bigg) + log \: \bigg(\dfrac{128}{125} \bigg )$

$\tt \implies \: log \: \bigg( \dfrac{25 \times 5}{64 \times 2} \bigg) + log \: \bigg( \dfrac{128}{125} \bigg) \qquad \qquad \pmb{ \bigg \lgroup \because \: log \: a + log \: b = log \: (a \times b) \bigg \rgroup}$

$\tt \implies \: log \: \bigg( \dfrac{125}{128 } \bigg ) + log \: \bigg( \dfrac{128}{125} \bigg)$

$\tt \implies \: log \: \bigg( \dfrac{125}{128} \times \dfrac{128}{125} \bigg) \qquad \qquad \pmb{ \bigg \lgroup\because \: log \: a \: + log \: b = log \: (a \times b) \bigg \rgroup }$

$\tt\implies \: log \: \bigg ( \cancel \dfrac{125}{128} \times \cancel\dfrac{128}{125} \bigg)$

$\tt \implies \: log \: (1) = 0 \qquad \qquad \pmb{ \bigg \lgroup \because \: log \: (1) = 0 \bigg \rgroup}$

$\tt \implies \: 0 = RHS$

$\tt \implies \: LhS = RHS$

$~$

Henceforth, proved!

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