Solving a Exponential equation How can I solve for n in the following equation:
$10 = 2^n + 3^n$
Thank you for the assistance.
 A: If $n\in\mathbb{N}$, there is no such $n$.
If $n\in\mathbb{R}$, there exists such $n$ by the continuity of function $f(x):=2^x+3^x$ and the use of https://en.wikipedia.org/wiki/Intermediate_value_theorem
A: Note: I will assume that $n\in \mathbb{R}$, since there does not exist a solution for $n\in \mathbb{N}$.
There does not exist a closed form solution for $n$ in terms of elementary functions. However, you can solve this numerically.
I will use the Newton-Raphson method.
Note 2: From now on, I will use $x$ instead of $n$ as the variable to find, just for convenience of the method.
$$2^x+3^x=10$$
The process is as follows:
$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} \tag{1}$$
The function $f$ to consider is:
$$f(x)=2^x+3^x-10$$
Evaluating its derivative:
$$f'(x)=2^x\ln(2)+3^x\ln(3)$$
Substituting into equation $(1)$:
$$x_{n+1}=x_n-\frac{2^{x_n}+3^{x_n}-10}{2^{x_n}\ln(2)+3^{x_n}\ln(3)}\tag{2}$$
We can iterate $(2)$ by using a spreadsheet or using more sophisticated software such as MATLAB. Let's start with a reasonable guess for the solution $x_0=2$.
$$\begin{array}{c|c}n&x_n\\\hline0&2\\1&1.76304\\2&1.72982\\3&1.72926\\4&1.72926\\5&1.72926\end{array}$$
Note that as the iterations $n\to \infty$, $x_n\to x$. Doing this iteration repeatedly gives:
$$x\approx 1.729255558981860$$
The function is strictly increasing for all $x\in \mathbb{R}$, therefore we know that there exists only one root.
