How do you condense $log (6 x) + log (4 x)$ $log(6x)+log(4x)$ ?

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Answer 1 · 2016-03-20T00:50:49

Using the product rule of logarithms ${log}_{a} x + {log}_{a} y = {log}_{a} (x y)$ $log_a x + log_a y = log_a (xy)$
we have that

$log 6 x + log 4 x = log 24 x^{2}$ $log 6x+log 4x=log 24x^2$

How do you condense log(6x)+log(4x)log(6x)+log(4x)log(6x)+log(4x)?