How do you solve $log4x - log16 = 1$ ?

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Answer 1 · 2016-04-14T22:33:36

Use the log rule $log_an - log_am = log_a(n/m)$

$log_10((4x)/16) = 1$

$(4x)/16 = 10^1$

$4x = 10(16)$

$4x = 160$

$x = 40$

Hopefully this helps!

How do you solve log4x - log16 = 1log4x - log16 = 1?