How do I solve $2^x + 3^x = 5$ type exponential equations?

Question

I was trying to solve a variety of exponential equations and I came across such equations and I am unable to solve it by the regular "taking log" method, so how do I solve such equations? (Well I know we can use hit and trial method and it comes out to $1$). But is there any proper defined method/algorithm that is used to solve such equations?

Similar type equations:

• $2^x + 3^x + 5^x - 10 = 0$

• $2^{2x} - 4^x$

Answer 1

Once you have identified an obvious solution, you can prove that there are no others by studying the sign of the derivative. Here $$f'(x)=\ln(2)2^x+\ln(3)3^x$$ is positive on $\Bbb R$, therefore there is only one solution, $x=1$ as you found.

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When the solution is not obvious, there is in general no systematic way to find one. You have to study it case by case and possibly do a numerical approximation.

Answer 2

In general - you don't. You use numerical methods.

Consider $2^x + 3^x = 6$, very close to your equation. It still has a solution, but there is likely no closed form for it.

Answer 3

As said in comments and answers, most of the time, numerical methods would be required.

Let us consider the case of $$\sum_{i=1}^n a_i^x=b$$ with $0 < a_1 <a_2 <\cdots < a_n$ and write the equation as $$a_n^x=b-\sum_{i=1}^{n-1} a_i^x$$ Take logarithms $$x \log(a_n)=\log\left(b-\sum_{i=1}^{n-1} a_i^x \right)$$ which will make the problem much more linear.

To get an estimate, start at $x=0$ and use a Taylor expansion of the rhs. This would make $$x \log(a_n)=\log(b-n-1)-\frac {\sum_{i=1}^{n-1} \log(a_i)}{b-n-1} x + O(x^2)$$ Ignoring the higher order terms, this will give an (over)estimate which can be used as the $x_0$ in Newton method in fact $x_0$ is the result of the first iteration of Newton method starting at $0$).

Your case $2^x + 3^x + 5^x - 10 = 0$ being trivial $(x=1)$, let us try changing $b=10$ to $b=100$. The estimate would be $x_0=2.81680$ and Newton iterates would be $$\left( \begin{array}{cc} k & x_k \\ 0 & 2.816800000 \\ 1 & 2.681921995 \\ 2 & 2.679370802 \\ 3 & 2.679370026 \end{array} \right)$$ $$ which is the solution for ten significant figures.

Let us repeat with $b=1000$; the estimate is $x_0=4.28600$ and Newton iterates would be $$\left( \begin{array}{cc} k & x_k \\ 0 & 4.286000000 \\ 1 & 4.212332825 \\ 2 & 4.212049819 \\ 3 & 4.212049815 \end{array} \right)$$ which is the solution for ten significant figures.