Solve equation $4^x-3\cdot6^x+2\cdot9^x=0$
$4^x-3\cdot6^x+2\cdot9^x=0, x\in\mathbb{R}$
Can someone solve this and explain me step by step cause I have no idea what to do.
Solution 1:
HINT:
See Exponent Combination Laws
Let $2^x=a,3^x=b$
$\implies 4^x=(2^x)^2=a^2,6^x=ab,9^x=b^2$
Divide both sides of the given equation by $9^x(=b^2)$
Finally Now if $\displaystyle u^m=1,$
either $\displaystyle m=0,u\ne0; $
or $\displaystyle u=1$
or $\displaystyle u=-1,m$ is even