Question

Solve the following equations, having given $\log 2,\log 3$, and $\log 7$.
$2^x.6^{x-1}=5^{2x}.7^{1-x}$.

Answer 1

Product Rule: ${\log }_{a}mn={\log }_{a}m+{\log }_{a}n$

Division Rule: ${\log }_a\frac{m}{n}={\log }_{a}m-{\log }_{a}n$

Power Rule: ${\log }_a{m}^n=n{\log }_{a}m$

$=>2^x.(2.3{)}^{x-1}=(\frac{10}{2}{)}^{2x}{.7}^{1-x}$

$=>2^{2x-1}{.3}^{x-1}={10}^{2x}{.2}^{-2x}{.7}^{1-x}$

Applying log on both sides,and by the above mentioned rules

$=>(2x-1)log2+(x-1)log3=2xlog10-2xlog2+(1-x)log7$

$=>x(4log2+log3+log7-2)=log3+log7+log2$

$=>x=\frac{log2+log3+log7}{4log2+log3+log7-2}$