I need help with this following exponential equation

Question

$9^x-6^x+4^x-3^x-2^x+1=0$
I have to find the number of real values of $x$ that satisfy the given equation.

Answer 1

With some experimentation, we can factor as follows: $$(3^x-2^x-1)3^x + (2^x-1)2^x + 1 = 0.$$

For $x>1$, the LHS will certainly be be positive, which doesn't agree with the RHS.

So we can at least rule out all $x>1$. (Not that one solution is $x = 0$).

Answer 2

Since $$9^x -6^x+4^x - 3^x-2^x+1 = (3^x-\frac{1}{2}\cdot 2^x - \frac{1}{2})^2 + \frac{3}{4}(2^x-1)^2 = 0$$

We have $3^x-\frac{1}{2}\cdot 2^x - \frac{1}{2} = 0, 2^x-1 = 0$ and thus $x = 0$.

Answer 3

Yet another approach is to rewrite LHS of a given equation as:
$(3^x-1)^2 - (3^x-1)(2^x-1)+(2^x-1)^2$

Then we can substitute $3^x-1$ with $u$, and $2^x-1$ with $v$, hence we will have:
$u^2 - uv + v^2 =0$.

Above equation is satisfied if and only if $u=v=0 \iff \boxed{x=0}$.