# Exponential Equation with different bases and a constant

Anyone know how one could algebraically solve an exponential equation with form:

$$0=a(b^x)+(c^{x-1})+d$$

My specific example is:

$$0=33.1(1.04^n)+(0.75^{n-1})-37.5$$

I tried getting rid of the "$$n-1$$" for:

$$0=24.825(1.04^n)+(0.75^n)-26.775$$

But I'm still stuck from there.

I know what the answer should be for graphing it, but I can't use a graphing calculator on the test. Thank you very much! (Have spent about 3 futile research hours tying to figure it out!)

First of all, I think that you made a small mistake since you should have arrived at $$24.825\, (1.04)^n+(0.75)^n-\color{red}{28.125}=0$$
Now, the problem is that you will not be able to get an anlytical solution of this equation and you will need a numerical method to find the zero of function $$f(n)=24.825\, (1.04)^n+(0.75)^n-28.125$$
By inspection, you should find that the solution is between $$n=2$$ (since $$f(2)=-0.712$$) and $$n=3$$ (since $$f(3)=0.222$$). At this point, you have all elements to start either bisection method or Newton method.
Using Newton method and using $$n_0=2.5$$, you should get the following iterates $$\left( \begin{array}{cc} k & n_k \\ 0 & 2.50000 \\ 1 & 2.77349 \\ 2 & 2.77030 \\ 3 & 2.77030 \end{array} \right)$$