High school math student here. In my homework I was asked to solve $16^x +4^{x+1} - 3= 0$ and I used substitution to get $x=\log_4{(-2+\sqrt7)}$. However, this was in the chapter on logarithms. How can I solve this without recourse to the quadratic formula etc, that is, using only logarithms and their properties and exponential functions and their properties? If I can't, how would one prove this is impossible?

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    $\begingroup$ Unfortunately, the way the equation is written does not permit naive use of logarithms because logarithms do not play nicely with addition. The quadratic equation is necessary here as far as I am concerned. $\endgroup$ – Cameron Williams Dec 2 '14 at 5:08
  • $\begingroup$ @CameronWilliams That's what I think too, but how do we prove it? $\endgroup$ – Elliot Gorokhovsky Dec 2 '14 at 5:09
  • $\begingroup$ It's not enough to use the properties of logarithms and exponentials--you obviously must also use properties of addition and equality (among other things), otherwise you can't even say what the equation means. Why would other things that you are supposed to know already be off limits, then? $\endgroup$ – David K Dec 2 '14 at 5:41
  • $\begingroup$ Speculation: If the Quadratic Formula or equivalent is not to be used, perhaps there is a typo, and it was intended to be $16^x-4^{x+1}+3=0$. $\endgroup$ – André Nicolas Dec 2 '14 at 6:31
  • $\begingroup$ You are not allowed to use the quadratic formula, and factoring does not work. You could use completing the square! $\endgroup$ – Rory Daulton Dec 2 '14 at 12:20

(Hopefully someone still finds this useful three and a half years later!)

This is the best way I can think to do this problem without the quadratic formula, and I suspect it may be similar to the way in which you did it.

$16^x+4^{x+1}-3=0$ can be rewritten as $(4^x)^2+4(4^x)-3=0$, and we can make the substitution $u=4^x$ to arrive at $u^2+4u-3=0$. Unfortunately, that doesn't factor nicely, so the next best option (assuming we're not allowed to use the quadratic formula) is to complete the square, as Rory Daulton suggested in a comment. $$\begin{align}u^2+4u-3&=u^2+4u+4-7\\&=(u+2)^2-7\\(u+2)^2-7&=0\\(u+2)^2&=7\\u+2&=\pm\sqrt{7}\\u&=-2\pm\sqrt{7}\end{align}$$

Now we substitute back in: $$\begin{align}4^x&=-2\pm\sqrt{7}\\x&=\log_4(-2+\sqrt{7})\end{align}$$

  • $\begingroup$ Thanks! I'm in university now and just finished taking Galois theory, so of course now I understand that completing the square is equivalent to using the quadratic formula :) $\endgroup$ – Elliot Gorokhovsky Apr 27 '18 at 19:39
  • $\begingroup$ Have to admit I was a little surprised to see a response, but you're welcome! I wasn't sure at first what level to write my answer at, so I figured I'd go with something that would be accessible to any other high school math students who come along with similar questions. $\endgroup$ – Robert Howard Apr 27 '18 at 21:15
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    $\begingroup$ Haha I understand, it's refreshing to see an algebra problem where I don't have to compute any permutation groups! $\endgroup$ – Elliot Gorokhovsky Apr 28 '18 at 20:23

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