We have:
$$16^x = 12^x + 9^x$$
Just by visual inspection one can say that the answer lies somewhere between $1$ and $2$.
I gave the starting point of the iteration as $2$ and plugged the function in and got the answer as $1.6727$. How do I prove that to be correct?
On
$16^x=12^x+9^x$
$9^x(16/9)^x=9^x(4/3)^x+9^x$
$9^x((16/9)^x-(4/3)^x-1)=0$
Divide both sides by $9^x$ since that quantity can never be zero.
$(16/9)^x-(4/3)^x-1=0$
$((4/3)^x)^2-(4/3)^x-1=0$
$(4/3)^x=\frac{1\pm\sqrt{5}}{2}$ (quadratic formula)
$x=\log_{4/3}\left(\frac{1\pm\sqrt{5}}{2}\right)$
$\frac{1-\sqrt{5}}{2}$ is negative and logs cannot have negative numbers in them, so I throw that solution out. The only solution is $x=\log_{4/3}\left(\frac{1+\sqrt{5}}{2}\right)$.
Start with:
$$16^x = 12^x + 9^x$$
Divide both sides by $12^x$
$$(4/3)^x = 1 + (3/4)^x$$
Multiply by $(4/3)^x$
$$((4/3)^x)^2 = (4/3)^x + 1$$
Substitute $a = (4/3)^x$ and solve using the quadratic formula. Remember to only keep the positive root.
$$a^2 - a - 1 = 0$$
$$a = \dfrac{1+\sqrt{5}}{2}$$
$$(4/3)^x = \dfrac{1+\sqrt{5}}{2}$$
Can you solve it from here?