How do you solve $log 4 x - log 16 = 1$ $log4x - log16 = 1$ ?

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Answer 1 · 2016-04-14T22:33:36

Use the log rule ${log}_{a} n - {log}_{a} m = {log}_{a} (\frac{n}{m})$ $log_an - log_am = log_a(n/m)$

${log}_{10} (\frac{4 x}{16}) = 1$ $log_10((4x)/16) = 1$

$\frac{4 x}{16} = 10^{1}$ $(4x)/16 = 10^1$

$4 x = 10 (16)$ $4x = 10(16)$

$4 x = 160$ $4x = 160$

$x = 40$ $x = 40$

Hopefully this helps!

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