Solve $8^{x-2} - 6^x = 56$ for $x$

Question

I was recently challenged to solve the following equation $8^{x-2} - 6^x = 56$ for $x$. There obviously has to be a solution, but how would you calculate that? I tried rewriting the left side from $a^2-b^2$ to $(a+b)(a-b)$ but that didn't really help me.

Answer 1

$y = x-2\\ 8^y - 36\cdot 6^y = 56\\ 6^y(\frac {8}{6}^y - 36) = 56\\ \frac {8}{6}^y - 36 > 0\\ y>\frac {\ln{36}}{\ln8-\ln6} = 12.456525$

Now we can substitue this number for $y$ to more tightly pin it down, if we felt it would to any good.

$\frac {56}{6^{13}}<\frac {8}{6}^y - 36 < \frac {56}{6^{12}}\\ \frac {\ln{36}+\frac {56}{6^{13}}}{\ln8-\ln6} < y <\frac {\ln{36+\frac {56}{6^{12}}}}{\ln8-\ln6} $

$x\approx \frac {\ln{36}}{\ln8-\ln6}+2$ is accurate to within $10^{-8}$

Answer 2

As already said, you need to use a numerical method for solving $$8^{x-2} - 6^x = 56\implies \frac{8^x}{64}-6^x=56\implies 8^x=64(56+6^x)$$ that is to say to find the zero of $$f(x)=8^x-64(56+6^x)$$ This function varies extremely fast (try to graph it) and it is better to consider $$g(x)=x\log(8)-\log(64(56+6^x))$$ Just by curiosity, make a graph of this function to see how nice how nice it is.

Let us try Newton method; being very lazy, let us start with $x_0=0$. The iterates are $$\left( \begin{array}{cc} n & x_n \\ 0 & 0 \\ 1 & 4.0048367463046664599 \\ 2 & 12.443399536487988071 \\ 3 & 14.456524932583593226 \\ 4 & 14.456525039015510556 \end{array} \right)$$ which is the solution for twenty significant figures.

Being less lazy, you could notice that $(8^6-6^8)$ is a huge negative number. So, start using $x_0=8$; the first iterate of Newton method is $14.4553$ and the second one is almost the solution for ten significant figures.