Question

Find x and y using Cramer's Rule if :$\frac{1}{2^x}+\frac{2}{3^y}=10,\frac{3}{2^x}-\frac{5}{3^y}=-3$ ?

Answer 1

$\frac{1}{2^x}+\frac{2}{3^y}=10⟶\left(1\right)\frac{3}{2^x}-\frac{5}{3^y}=-3⟶\left(2\right)$

Let $2^x=p$ and $3^y=q$

Equation becomes;

$\frac{1}{p}+\frac{2}{q}=10⟶\left(3\right)\frac{3}{p}-\frac{5}{q}=-3⟶\left(4\right)$

again let, $\frac{1}{p}=u$ and $\frac{1}{q}=v$

Equation (3) and (4) becomes

$u+2v=10⟶\left(5\right)3u-5v=-3⟶\left(6\right)$

By Cramer's rule;

$u=\frac{\left|\begin{array}{cc}{c}_1& b_1\\ c_2& b_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}=\frac{\left|\begin{array}{cc}10& 2\\ -3& -5\end{array}\right|}{\left|\begin{array}{cc}1& 2\\ 3& -5\end{array}\right|}=4$

and,

$v=\frac{\left|\begin{array}{cc}{a}_1& c_1\\ a_2& c_2\end{array}\right|}{\left|\begin{array}{cc}{a}_1& b_1\\ a_2& b_2\end{array}\right|}=\frac{\left|\begin{array}{cc}1& 10\\ 3& -3\end{array}\right|}{\left|\begin{array}{cc}1& 2\\ 3& -5\end{array}\right|}=3$

$\because \frac{1}{p}=u;\frac{1}{q}=v\Rightarrow p=\frac{1}{4};q=\frac{1}{3}$

again,

$2^x=p;3^y=q\Rightarrow 2^x=\frac{1}{4};3^y=\frac{1}{3}\Rightarrow x\log 2=\log \frac{1}{4};y\log 3=\log \frac{1}{3}\Rightarrow x=\log \frac{1}{8};y=\log \frac{1}{9}$