A=P(index0)10^r

If A=80, P(index0)=90, find r

Here's what I did:

A=P(index0)10^r

80=90^r

ln80=ln90^r

r=ln80/ln90

so I get r= 0.973 but actually the solution is supposed to be r=log(8/9) or -0.051153.

What did I do wrong?

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- Jul 31st 2008, 09:07 AMtami-chan87solving an exponential equation
A=P(index0)10^r

If A=80, P(index0)=90, find r

Here's what I did:

A=P(index0)10^r

80=90^r

ln80=ln90^r

r=ln80/ln90

so I get r= 0.973 but actually the solution is supposed to be r=log(8/9) or -0.051153.

What did I do wrong? - Jul 31st 2008, 09:30 AMChris L T521
You forgot that r was the exponent of 10, not 90.

Thus, $\displaystyle \tfrac{80}{90}=10^r$

Since $\displaystyle \log (x) = \log_{10} (x)$, we can take the logs of both sides:

$\displaystyle \log\left(\tfrac{8}{9}\right)=\log\left(10^r\right )\implies \color{red}\boxed{r=\log\left(\tfrac{8}{9}\right)}$

Does this make sense?

--Chris - Jul 31st 2008, 09:53 AMtami-chan87
Yeah, thanks for your help :)