Find all real $a$ such that $6a^2+3=9^a$

Question

Find all real $a$ such that $6a^2+3=9^a$
The problem seems to be very easy, but now i can't see an easy way to find if there are other roots than $1$. Tried using the derivative but that didn't help me.
Thanks for your answers.

Answer 1

Possible solution outline:

Set $f(a)= 6a^2+3-9^a$.

Then, as you observed, $f(1) = 0$. To deduce that this is the only root, try to show the that $f(a)$ is monotonically decreasing, i.e. $f'(a) < 0$ for all $a$. To do this, find the point for the maximum of $f'(a)$ by solving $f''(a)=0$.

$$f''(a) = 12-9^a\log^29 = 0\\\implies \\a=\frac{\log\left(\frac{12}{\log^2 9}\right)}9$$

(To show it's indeed a maximum, we observe that $f'''(a)=-9^a\log^39 <0$ for all $a$.)

Plugging in the solution into $f'(a)$ gives us a negative value, and we can conlude that $f(a)<0$ for all $a$, hence $a=1$ is the only solution to $f(a)=0$.

Answer 2

Try proving that for $a<1$ the left side is greater (obvious for negative $a$ and positive where $9^a<3$). For $a>1$ check that the right side is greater (the derivative would tell you which side is steeper).