Equation concerning different bases and exponents

Question

I have an equation that I've been trying to solve for fun algebraically after helping someone learn logarithm and exponent rules: $$3^x+11^{2x^2+4x-7}=311$$ So far, no manipulation using logarithms or the changing of bases has helped. I know it is probably not possible to solve it like this, but I was wondering if there is any method to getting an exact solution, not a numerical one (Desmos gives ~-3.3868 and ~1.38616). If there is no method, how come? I feel like there should be some expression that perfectly captures the solution, whether it be algebraic, complex, or something else. My goal is to figure out how to solve these more difficult exponential equations in case someone asks me. Sorry if this question is relatively simple compared to others on this site, but I don't know where to ask this.

Answer 1

There is solution as an infinite series.

$$f(x)=3^x+11^{2x^2+4x-7}-311$$

I assume that you noticed thet $f(x)$ is very stiff and that the positive solution is rather close to $x=\frac 32$.

So, consider instead a smooth transform of $f(x)$ such as $$g(x)=(2x^2+4x-7)\log(11)-\log(311-3^x)$$ and expand it as a Taylor series around $x=\frac 32$.

This gives $$g_p(x)=\sum_{n=0}^p a_n\,\left(x-\frac{3}{2}\right)^n+O\left(\left(x-\frac{3}{2}\right)^{p+1}\right)$$ The $a_n$ as quite nasty but this does not matter.

Now, use power series reversion to get explicitly $x_{(p)}$ using the explicit formula given by Morse and Feshbach for the $n^{\text{th}}$ term of the inverse series.

For example $$x_{(1)}=\frac 32-\frac{48347 \left(7 \log (11)-2 \log \left(311-3 \sqrt{3}\right)\right)}{27 \log (3)+933 \sqrt{3} \log (3)+966940 \log (11)}=1.38875$$ while the solution is $1.38616$.

Playing with $p$ and converting the result to decimals $$\left( \begin{array}{cc} p & x_{(p)} \\ 1 & 1.38875190259 \\ 2 & 1.38627322206 \\ 3 & 1.38616299535 \\ 4 & 1.38615686078 \\ 5 & 1.38615647853 \\ 6 & 1.38615645301 \\ 7 & 1.38615645122 \\ 8 & 1.38615645109 \\ 9 & 1.38615645108 \\ \end{array} \right)$$

and we could continue forever.

For example $x_{(15)}=\color{red}{1.38615645108209659}12$

For the negative root, do the same around $x=-\frac 72$; for this case $$x_{(1)}=-\frac 72+\frac{\left(\sqrt{3}-25191\right) \left(8 \log (3)+7 \log (11)-2 \log \left(25191-\sqrt{3}\right)\right)}{2 \left(\sqrt{3} \log (3)-251910 \log (11)+10 \sqrt{3} \log (11)\right)}$$ which is $-3.38936$ while the solution is $-3.38680$.

For example $x_{(15)}=\color{red}{-3.386801949501781337}46$