If $\sqrt{9^x - 6^x} = \sqrt{6^x - 4^x}$, then the number of values of $x$ are:

Question

If $\sqrt{9^x - 6^x} = \sqrt{6^x - 4^x}$, then the number of values of $x$ are:

I have solved the above like this:

$\sqrt{9^x - 6^x} = \sqrt{6^x - 4^x}$

Squaring:

$9^x - 6^x = 6^x - 4^x$

$9^x + 4^x = 2\times(6^x)$

$3^{2x} + 2^{2x} = 2\times (3\times2)^{x}$

$(3^{x} - 2^{x})^{2} = 0$

Giving us, $3^x = 2^x$, so $x=0$ is the only value which satisfies.

But in the answers, they have stated that total solutions are $3$. Where have I made a mistake? And at what step did a specific action lead to loss of roots?

Please help me to go about the same. Thanks!

Answer 1

Your answer is correct. Here is a slightly different way of getting the same answer:\begin{align}9^x-6^x=6^x-4^x&\iff\left(\frac32\right)^{2x}-2\left(\frac32\right)^x+1=0\\&\iff\left(\left(\frac32\right)^x-1\right)^2=0\\&\iff\left(\frac32\right)^x=1\\&\iff x=0.\end{align}

Answer 2

Factorize the equation

$$\sqrt{9^x - 6^x} - \sqrt{6^x - 4^x} =\sqrt{3^x(3^x - 2^x)}- \sqrt{2^x(3^x - 2^x)}\\ = \sqrt{3^x - 2^x}\left(\sqrt{3^x}-\sqrt{2^x} \right)=0 $$