How do you solve the equation $log_2n=1/4log_2 16+1/2log_2 49$ ?

Question

2900 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-30T02:58:49

$n=14$

$log_2 n =1/4 log_2 16 + 1/2 log_2 49$

First use the log rule $alogx=logx^a$ .

$log_2 n =log_2 16^(1/4)+ log_2 49^(1/2)$

$16^(1/4)=root(4)16=2$ and $49^(1/2)=sqrt49=7$

$log_2 n = log_2 2 +log_2 7$

Use the log rule $logx +logy = logxy$ to condense the log.

$log_2 n = log_2 (2 *7)$

$log_2 n = log_2 14$

$n=14$

How do you solve the equation log_2n=1/4log_2 16+1/2log_2 49log_2n=1/4log_2 16+1/2log_2 49?