# Solve all for all real $x$: $9^x +6^x = 4(4^x)$

I am having some trouble finding all real solutions to $$9^x +6^x = 4(4^x)$$. The original equation given was $$3^x(3^x + 2^x) = 4^{x+1}$$. I've fiddled around with it and tried to make everything the same base, but I'm not really sure what to do. Could someone tell me where to go next? (I'm pretty sure $$0$$ is the only real solution, so any hints on how to prove this would be greatly appreciated, or any corrections if I'm incorrect in this belief)

Note: For those people who are suspicious of everything - I've been on math stack exchange for quite a while (as well as accounts in other communities) , but I only just created an account for this :). This question comes from some extension work my maths teacher gave me (I'm scared to ask her for help haha).

Thank you!

• If the original equation was $3^x(3^x \cdot 2^x) = 4^{x + 1}$, shouldn't you have $9^x \cdot 6^x = 4 \cdot 4^x$? Commented Feb 27, 2022 at 10:09
• "Please don't suggest using something overcomplicated such as log" The answer is a logarithm. You might as well say, "don't use $\sqrt{}$ to solve $x^2-17=0$".
– J.G.
Commented Feb 27, 2022 at 10:13
• What I am questioning is how you got an addition sign on the left-hand side. Commented Feb 27, 2022 at 10:21
• Oh sorry! I typed that wrong oopsy. I'll go edit it now.
– user1030084
Commented Feb 27, 2022 at 10:26
• Observe that $0$ cannot be a solution since $9^0 + 6^0 = 1 + 1 = 2 \neq 4 = 4 \cdot 1 = 4 \cdot 4^0$. Commented Feb 27, 2022 at 10:44

Hint...divide by $$4^x$$ and substitute $$u=(\frac32)^x$$

The zero of function $$f(x)=9^x +6^x - 4(4^x)$$ is not $$x=1$$ since $$f(1)=-1$$ but it should close.

Consider instead $$g(x)=\log[9^x +6^x]-\log[4(4^x)]$$ which is almost a straight line. Expand it as Taylor series $$g(x)=-\log \left(\frac{16}{15}\right)+\frac{8}{5} \log \left(\frac{3}{2}\right)(x-1)+O\left((x-1)^2\right)$$ which gives, as an estimate, $$x \sim 1+\frac{5 \log \left(\frac{16}{15}\right)}{8 \log \left(\frac{3}{2}\right)}=1.09948$$ while the solution given by Newton method is $$x=1.09918$$.

After a few manipulations you can see the exact solution is $$x =\frac{log(W)}{log (3) -log (2) }$$

where $$W= \frac{\sqrt{17}-1}{2}$$ and log representa the natural logarithm.

$$x=1.099183900415395$$