Question

$proved \: that \: log_{2}(10) - log_{5}(125) \times log_{8}(5) = 1$
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Answer 1

Step-by-step explanation:

LHS:

$log_{2}(10) - log_{5}(125 ) \times log_{8}(5) \\ = \frac{ log(10) }{ log(2) } - \frac{ 3log(5) }{ log(8) } \\ = \frac{1}{ log(2) } - \frac{3 log(5) }{3 log(2) } \\ = \frac{1 - log(5) }{ log(2) } \\ = \frac{1 - 0.6989}{0.3011} \\ = \frac{0.3011}{0.3011} \\ = 1$

I have taken the approx values of log2 and log5(for simplicity), but even if you take full value the result won't change.

now LHS = RHS hence prooved

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