Question

Draw the exponential which passes through the points (-1, 8) and (3, 3). From your graph, estimate the
halving time from your graph. Find a formula for the exponential, and estimate the halving time from
your formula by guessing and checking. Using logarithms, nd the exact halving time and compare this to
your estimate.

Answer 1

So far by doing y=m*x+b and using \[\frac{ 3-8 }{ 3-(-1) } = \frac{ -5 }{4} = m\]

Answer 2

Then working out out:
8 = -5/4*(-1)+b =
5/4 + 27/4

Thus m = -5/4
b = 27/4

Is this all corret so far?

Answer 3

Just found also using y= a(b^x), Which I assume is how I proceed?

Answer 4

Basically, you have two equations and two variables: \[
8=ab^{-1} \\
3=ab^3
\]

Answer 5

Solve for \(a\) in terms of \(b\), then substitute it into the other equation.

Answer 6

If I multiply 8*b^3 and 3*b^-1 will this help find the b value?

Answer 7

I am using this site http://www.ehow.com/how_8117999_exponential-equation-two-points.html but it is confusing on how to proceed.

Answer 8

Okay first of all in the first equation we solve for \(a\) \[
8=ab^{-1}\implies 8b=a
\]Then we substitute in to the second equation \[
3 = ab^3 = (8b)b^3 = 8b^4
\]

Answer 9

This means \[
b = \sqrt[4]{\frac{3}{8}}
\]

Answer 10

@Astrobuoy Do you see what I did there?

Answer 11

thus \[8*\sqrt[4]{\frac{ 3 }{ 8 }} = a\]

Hence:

\[4*\sqrt[4]{6} = a\]

Answer 12

Hmm, now did you get that?

Answer 13

\[
8\sqrt[4]{\frac{3}{8}}=\sqrt[4]{\frac{8^4}{1}}\sqrt[4]{\frac{3}{8}} = \sqrt[4]{3\cdot 8^3}
\]

Answer 14

You wrote eariler 8b = a, So I mutiplied 8 by the b that is listed there.

Answer 15

Strange, I did the problem on Wolfram and it came up as what I wrote. :\

Answer 16

It is strange.

Answer 17

Hmm, maybe they did \(8=4\cdot 2=4\sqrt[4]{16}\)

Answer 18

Perhaps a syntax error on mine or the systems behalf. So my final a = \[\sqrt[4]{3*8^{3}} ?\]

Answer 19

no, I think actually it is correct, I just didn't simply mine enough.

Answer 20

http://www.wolframalpha.com/input/?i=8*%283%2F8%29%5E%281%2F4%29%29

Is the link from my page of computation.

Answer 21

anyway we we done?

Answer 22

cheers, I can take it from here :) thanks for the help!