How do you solve $3 {log}_{4} n = 6$ $3log_4 n= 6$ ?

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Answer 1 · 2016-04-04T23:10:17

$n = 16$ $n = 16$

Begin with dividing both sides by three to make it much easier to approach.

$4 {log}_{4} n = 6$ $4log_4n = 6$
${log}_{4} n = 2$ $log_4n = 2$

Now raise both sides to the power of 4, which will cancel out the logarithm

${log}_{4} n = 2$ $log_4n = 2$
$4^{{log}_{4} n} = 4^{2}$ $4^(log_4n) = 4^2$
$n = 4^{2}$ $n = 4^2$
$n = 16$ $n = 16$

How do you solve 3log_4 n= 6 3log4n=63log_4 n= 6 ?