Question

Type the correct answer in the box. Log14/13+log11/5-log22/15= log

Answer 1

$\text{Use}\\\\\log_ab+\log_ac=\log_a(bc)\\\\\log_ab-\log_ac=\log_a(b:c)\\\\\tet{for}\ a>0\ \wedge\ a\neq1\ \wedge\ b>0\ \wedge\ c>0\\-----------------------------\\\\\log\dfrac{14}{13}+\log\dfrac{11}{5}-\log\dfrac{22}{15}=\log\left(\dfrac{14}{13}\cdot\dfrac{11}{5}:\dfrac{22}{15}\right)=\log\left(\dfrac{14}{13}\cdot\dfrac{11\!\!\!\!\!\diagup^1}{5\!\!\!\!\diagup_1}\cdot\dfrac{15\!\!\!\!\!\diagup^3}{22\!\!\!\!\!\diagup_2}\right)$

$=\log\left(\dfrac{14\!\!\!\!\!\diagup^7}{13}\cdot\dfrac{1}{1}\cdot\dfrac{3}{2\!\!\!\!\diagup_1}\right)=\log\dfrac{21}{13}$

Answer 2

your answer would be x=21/13

When the logs have the same base: logb(f(x)) = logb(g(x)) --> f(x) g(x)

if im correct please mark brainliest :)