Solve equation $4^x-3\cdot6^x+2\cdot9^x=0$

Question

$4^x-3\cdot6^x+2\cdot9^x=0, x\in\mathbb{R}$

Can someone solve this and explain me step by step cause I have no idea what to do.

Answer 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

Answer 2

$$4^x-3\cdot6^x+2\cdot9^x=0$$ $$(2^x)^2-3\cdot 2^x\cdot3^x+2\cdot(3^x)^2=0$$ $$\left(\frac23\right)^{2x}-3\cdot\left(\frac23\right)^{x}+2=0$$ Let $ \left(\frac23 \right)^{x}=t$

$$t^2-3t+2=0$$

$\left(\frac23\right)^{x}=1$ or $\left(\frac23\right)^{x}=2$

$x=0$ or $x=\log_{\frac23}2$

Answer 3

Hint:

write the equation as: $$ 2^{2x}-3\cdot 2^x\cdot 3^x+2\cdot 3^{2x}=0 \quad \iff \quad 2^{2x}-2\cdot 2^x\cdot 3^x-1\cdot 2^x\cdot 3^x+2\cdot 3^{2x}=0 $$ that can be factorized as: $$ \left(2^x-2\cdot 3^x \right)\left(2^x-3^x \right)=0 $$

Now you can solve: $$ \left(2^x-2\cdot 3^x \right)=0 \quad \lor \quad \left(2^x-3^x \right)=0 $$