How to solve $3 + 0.075x = 1.075^x$ for x?

Question

I tried various methods... Could only solve it numerically not analytically. Any suggestions?

Answer 1

The equation $$a+b x=c^x$$ has solutions in terms of Lambert function. It is given by $$x=-\frac{1}{\log (c)}W\left(-\frac{\log (c) }{b}c^{-\frac{a}{b}}\right)-\frac{a}{b}$$ but this function has multiple branches.

Consider that you look for the zero's of $$f(x)=c^x-bx-a$$ which has as derivatives $$f'(x)=c^x \log (c)-b\qquad \text{and} \qquad f''(x)=c^x \log ^2(c)\, \, >0 \, \forall x$$ The first derivative cancels at $$x_*=\frac{\log \left(\frac{b}{\log (c)}\right)}{\log (c)}$$ which is aminimum. So, if $f(x_*)<0$ tow possible roots. If you want to approximate them, develop as Taylor series aound this point $$f(x)=f(x_*)+\frac{1}{2} (x-x_*)^2 f''(x_*)+O\left((x-x_*)^3\right)$$ and solve the quadratic to get as estimates $$x_\pm=x_* \pm \sqrt{-2\frac{f(x_*)}{f''(x_*)}}$$

Applied to your case, this would give $x_-=-26.6577$ and $x_+=27.6638$; using them as starting values, Newton method would converge very fast to the solutions.

Answer 2

You can solve it with MatDeck as well, the equations solutions are -39.218 and 21.033

Solving of nonlinear equations in MatDeck