How do you condense $2 log 5 x - log 5 y$ $2 Log5 x - log5 y$ ?

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Answer 1 · 2016-04-11T18:18:16

$log (\frac{x^{2}}{y})$ $log(x^2/y)$

Condense the first $log$ $log$ , using the law that a coefficient outside a logarithm is the same as a power inside.

$2 log 5 x = log 5 x^{2}$ $2log5x = log5x^2$

Now condense everything, knowing that $log a - log b = log (\frac{a}{b})$ $loga - logb = log(a/b)$ .

$log 5 x^{2} - log 5 y = log (\frac{5 x^{2}}{5 y}) = log (\frac{x^{2}}{y})$ $log5x^2 - log5y = log((5x^2)/(5y)) = log(x^2/y)$

How do you condense 2log5x−log5y2 Log5 x - log5 y ?