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### Solving by Interpolation

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Date: 05/30/2001 at 05:04:47
From: Rosly
Subject: Exponential equation

Dear Dr. Math,

Given y = 10 and b = 1.419, find X from this equation:

y = ( b^(-0.25X)) + 0.84X

I have X, but when I replaced in the equation above, I couldn't get
y = 10.

```

```
Date: 05/30/2001 at 16:07:00
From: Doctor Rob
Subject: Re: Exponential equation

Thanks for writing to Ask Dr. Math, Rosly.

y = 1.419^(-0.25*X) + 0.84*X
= (1.419^[-0.25])^X + 0.84*X
y = 0.91623^X + 0.84*X

This equation, and many other similar ones, cannot be solved
algebraically. You will have to find the answer numerically.

One way to do this is by the method of interpolation. Find values of X
that make y < 10 and y > 10. Call them X1 and X2, respectively. The
actual value of X lies between them. Call the corresponding values of
y y1 and y2, respectively. Then compute a new value

X3 = X1 + (X2-X1)*(10-y1)/(y2-y1)

This is the y-intercept of the straight line through (X1,y1) and
(X2,y2). If the curve is straight, this will be the answer, and even
if it is not, it is closer to the answer than one of X1 or X2.
Compute the corresponding y3. Replace (X1,y1) by (X2,y2) and (X2,y2)
by (X3,y3). Repeat this until you get as much accuracy as you need.

For example,

X                 y
10                8.817
12               10.430
11.46687          9.99887
11.468267964      9.999998975
11.46826923329   10.00000000000

This is the answer to 11 decimal places. The actual value is about
11.46826923328992, according to Mathematica(TM).

You can see that this process converges quite rapidly to the

- Doctor Rob, The Math Forum
http://mathforum.org/dr.math/
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