Finding the number of roots of an equation

Question

Number of real roots of the equation $2^x + 2^{x-1} + 2^{x-2} = 7^x + 7^{x-1} + 7^{x-2} $ are

This is how i attempted it

$2^x + \frac {2^x}{2} + \frac {2^x}{2^2} = 7^x + \frac {7^x}{2} + \frac {7^x}{2^2} $

$= 2^x (1 + \frac {1}{2} + \frac {1}{2^2}) = 7^x(1 + \frac {1}{2} + \frac {1}{2^2})$

$2^x = 7^x$

$(\frac {2}{7})^x =1$ Therefore $x=0$ is the only solution. Is this right ? Or are there more solutions ?

Answer 1

Hint:$$2^x(1+\frac{1}{2}+\frac{1}{4})=7^x(1+\frac{1}{7}+\frac{1}{49})$$

So $x=0$ does not fit this equation.

$$\left(\frac{2}{7}\right)^x=\frac{228}{343}$$ You can solve for $x$ here by taking logs of both sides.