Question

Solve the following equations, having given $\log 2,\log 3$, and $\log 7$.
$\begin{array}{c}{2}^{x+y}=6^y\\ 3^x={3.2}^{y+1}\end{array}\}$

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

Given $2^{x+y}=6^y;3^x={3.2}^{y+1}$

$2^{x+y}=(2.3{)}^y$

$=>2^x=3^y$

Applying log on both sides

$xlog2=ylog3$

$=>y=\frac{xlog2}{log3}$

$3^x={3.2}^{y+1}$

$=>3^{x-1}=2^{y+1}$

Applying log on both sides

$(x-1)log3=(y+1)log2$

Substituting $y=\frac{xlog2}{log3}$

$(x-1)log3=(\frac{xlog2}{log3}+1)log2$

$=>x(log3{)}^2-(log3{)}^2=x(log2{)}^2+log2log3$

$=>x=\frac{log3(log3+log2)}{(log3+log2)(log3-log2)}$; Since$a^2-b^2=(a+b)(a-b)$

$=>x=\frac{log3}{log3-log2}$

$y=\frac{log2}{log3-log2}$