Question

Solve: $\frac{1}{3^x}+\frac{2}{2^y}=3$; $\frac{4}{3^x}+\frac{3}{2^y}-7=0$

Answer 1

A] $\frac{1}{3^x}+\frac{2}{2^y}=25---\left(1\right)$

$\frac{3}{3^x}+\frac{4}{2^y}=-5---\left(2\right)$

Let $1/3^x=a$ $1/2^y=b$

a+2b=25---(3)

3a+4b=-5---(4)

multiply (3) by 2 55+2b=25

2a+2b=50 2b=-30

3a+4b=-5 $\therefore$ b=-15

- - +

____________

-a = 55 Thus $\frac{1}{3a}=-55$

$\therefore$ a=-55

$3^x=\frac{-1}{55}\Rightarrow x=lo{g}_3\left(\frac{-1}{55}\right)$

$\frac{1}{2y=-15\Rightarrow y=lo{g}_2}\left(\frac{-1}{15}\right)$