How do you prove $ln (\frac{16}{3 x}) = 4 ln 2 - (ln 3 + ln x)$ $ln (16/(3x)) = 4ln2 - (ln3 + lnx)$ ?

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How do you prove $ln (\frac{16}{3 x}) = 4 ln 2 - (ln 3 + ln x)$ $ln (16/(3x)) = 4ln2 - (ln3 + lnx)$ ?

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Answer 1 · 2016-03-20T08:10:10

Lets Take RHS

$4 ln 2 - (ln 3 + ln x) = 4 ln 2 - ln 3 x$ $4ln2 - (ln3 + lnx) = 4 ln2 - ln3x$

$4 ln 2 - ln 3 x = {ln 2}^{4} - ln 3 x = ln 16 - ln 3 x = ln (\frac{16}{3} x) = L H S$ $4 ln2 - ln3x = ln 2^4 - ln3x = ln16 - ln 3x = ln(16/3x) = LHS$

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How do you prove $ln (\frac{16}{3 x}) = 4 ln 2 - (ln 3 + ln x)$ $ln (16/(3x)) = 4ln2 - (ln3 + lnx)$ ?

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How do you prove $ln (\frac{16}{3 x}) = 4 ln 2 - (ln 3 + ln x)$ $ln (16/(3x)) = 4ln2 - (ln3 + lnx)$ ?