How do you solve $4 {log}_{12} 2 + {log}_{12} x = {log}_{12} 96$ $4 log_12 2 + log_12 x = log_12 96$ ?

Question

3367 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-06T14:36:45

${6}$ ${6}$

Unite the logs.

${log}_{12} (2^{4} \cdot x) = {log}_{12} 96$ $log_12 (2^4 * x) = log_12 96$

$A = B \Rightarrow 12^{A} = 12^{B}$ $A = B Rightarrow 12^A = 12^B$

$16 \cdot x = 96$ $16 * x = 96$

$x = 6$ $x = 6$

How do you solve 4log122+log12x=log1296 4 log_12 2 + log_12 x = log_12 96?