Solve equation with exponentials in reals

Question

I have tried several ways to calculate the value of $x$ by analytical methods but I get nothing.

$$7.2^x = 16 + 5.3^x$$

Can someone help me?

Answer 1

I don't think there is as straightforward analytic solution, so use numerics. In Mathematica:

NSolve[7.2^x == 16 + 5.3^x, x, Reals]

$$x \to 1.83249$$

Here is a plot of $7.2^x - 16 - 5.3^x$:

Answer 2

Instead of considering that you look for the zero of function $$f(x)=7.2^x - 16 - 5.3^x$$ consider instead that you look for the zero of function $$g(x)=\log(7.2^x)-\log(16 + 5.3^x)=x \log(7.2)-\log(16 + 5.3^x)$$ which correspond to the intersection of a straight line and an almost straight line .

Taking this into account, to get an approximation of the solution, being myself quite lazy, use a Taylor expansion buile at $x=0$. This would give $$g(x)=x \log(7.2)-\log (17)-\frac{1}{17} x \log (5.3)+O\left(x^2\right)$$

This would give $$x_0=\frac{17 \log (17)}{17 \log \left(7.2\right)-\log \left(5.3\right)}\approx 1.51026$$

Now, start Newton method to get the following iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 1.51026 \\ 1 & 1.80366 \\ 2 & 1.83221 \\ 3 & 1.83249 \end{array} \right)$$

Changing the $16$ to $123456789$ would have given $$\left( \begin{array}{cc} n & x_n \\ 0 & 9.43801 \\ 1 & 9.46664 \\ 2 & 9.46667 \end{array} \right)$$