How can I find the value of the unknown X using any algebraic method? Not numerical methods and programming.

Question

$$72^x+7^x=80$$

The only clue that I have is that using limits is possible to find the unknown number (L'Hôpital's Rule)... but anyway, if you use other method, please let me know and I will appreciate it.

Answer 1

There is a unique real solution $x$, but it cannot be expressed in terms of elementary functions. Of course you can use numerical methods to approximate the value of $x$, and it should be clear that $x$ is very slightly larger than $1$. With the help of a computer I quickly found that $$x\approx1.0030899874071590958.$$

Answer 2

You can obtain approximations of the solution working around $x=1$ and using Taylor expansions to $O\big((x-1)^{n+1}\big)$ of $$y=72^x+7^x-80$$ For the lowest case $$y=-1+(x-1) (7 \log (7)+72 \log (72))+O\left((x-1)^2\right)$$ Now, using series reversion $$x=1+\frac{y+1}{7 \log (7)+72 \log (72)}+O\left((y+1)^2\right)$$ and make $y=0$.

It is sure that the expressions become very messy when we increase $n$.

As a function of $n$, the numerical value of the approximation would be $$\left( \begin{array}{cc} 1 & \color{blue}{1.003}110020086640054432582 \\ 2 & \color{blue}{1.003089}815152514185756329 \\ 3 & \color{blue}{1.00308998}9068733934362861 \\ 4 & \color{blue}{1.003089987}390094948474687 \\ 5 & \color{blue}{1.003089987407}341396343243 \\ 6 & \color{blue}{1.00308998740715}7094728794 \\ 7 & \color{blue}{1.003089987407159}118243290 \\ 8 & \color{blue}{1.003089987407159095}583690 \\ 9 & \color{blue}{1.0030899874071590958}41276 \\ 10 & \color{blue}{1.0030899874071590958383}13 \\ \cdots & \cdots \\ \infty & \color{blue}{1.003089987407159095838347} \end{array} \right)$$