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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!)

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1 Answer 1

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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 $$\left( \begin{array}{cc} k & n_k \\ 0 & 2.50000 \\ 1 & 2.77349 \\ 2 & 2.77030 \\ 3 & 2.77030 \end{array} \right)$$

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