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?

  • $\begingroup$ You can verify that it satisfies the equation. $\endgroup$ – M. Vinay Jun 9 '14 at 9:20
  • $\begingroup$ If the function changes sign between $1$ and $2$, then by the intermediate value theorem, it is guaranteed that the interval contains a root. Using any root finding method, you can narrow down the bracketing. $\endgroup$ – Yves Daoust Jun 9 '14 at 9:21

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?

  • $\begingroup$ Exactly what I needed. Thank you. $\endgroup$ – MadStacker Jun 9 '14 at 9:31
  • $\begingroup$ Elegant solution! +1 :) $\endgroup$ – Sergio Sarmiento Jun 9 '14 at 9:39
  • $\begingroup$ What is the answer you get @MadStacker? $\endgroup$ – Sergio Sarmiento Jun 9 '14 at 9:41




Divide both sides by $9^x$ since that quantity can never be zero.


$(4/3)^x=\frac{1\pm\sqrt{5}}{2}$ (quadratic formula)


$\frac{1-\sqrt{5}}{2}$ is negative and logs cannot have negative numbers in them, so I throw that solution out. The only solution is $x=\log_{4/3}\left(\frac{1+\sqrt{5}}{2}\right)$.


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