solving the equation by reducing it to quadratic form

Question

Solve for $x$ $$4^x-\dfrac{3^x}{\sqrt3}=3^x\cdot\sqrt3-\dfrac{2^{2x}}2$$ I don't understand how to convert it into quadratic equation how should I equate all bases

Answer 1

As $2^{2x}=(2^2)^x=4^x,$

$$4^x-\dfrac{3^x}{\sqrt3}=3^x\sqrt3-\dfrac{2^{2x}}2$$

$$2^{2x}\left(1+\dfrac12\right)=3^x\left(\sqrt3+\dfrac1{\sqrt3}\right)$$

$$2^{2x}\cdot\dfrac32=3^x\dfrac4{\sqrt3}$$

Take log in both sides to get $$2x\ln2+\ln3-\ln2=x\ln3+2\ln2-\dfrac12\ln3$$

$$\ln2(2x-1-2)=\ln3(x-\dfrac12-1)$$

$$\iff(2\ln2-\ln3)(2x-3)=0$$

Can you take it from here?