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Mathematics 21 Online
OpenStudy (anonymous):

The price of 9-volt batteries is increasing according to the function below, where t is years after January 1, 1980. During what year will the price reach $3? P(t) = 1.1 e0.047t

OpenStudy (anonymous):

do you mean \[P(t)=1.1e^{0.047t}\] or \[P(t)=1.1\times e \times 0.047t\]

OpenStudy (anonymous):

the first one sorry

OpenStudy (anonymous):

Ahh, I thought so Anyways you are given \[P(t)=3\] Therefore \[1.1e^{0.047t}=3\] \[e^{0.047t}=\frac{3}{1.1}\] Now, do you know how to solve after this for t?

OpenStudy (anonymous):

Not really :(

OpenStudy (anonymous):

I would divide 3 by 11

OpenStudy (anonymous):

Have you not read about logarithms?

OpenStudy (anonymous):

I have but I'm having hard time understanding it

OpenStudy (anonymous):

Ok, at this point you'd have to take log to the base e(also known as the natural log or just ln for short) on both sides of equation

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

Here are some rules, \[\log_{e}(e^x)=\ln(e^x)=x \] \[\log _{e}\frac{a}{b}=\ln\frac{a}{b}=\log_{e}(a)-\log_{e}(b)=\ln(a)-\ln(b)\] try using these rules to solve that equation, let me know if you have any problems

OpenStudy (anonymous):

Ok, Thank you

OpenStudy (anonymous):

\[\log_{e}(e)=1\] and \[\log_{e}(e^x)=xlog_{e}(e)=x\]

OpenStudy (anonymous):

the answer I got was 21.35...

OpenStudy (anonymous):

let me check

OpenStudy (anonymous):

\[T = \ln (3/1.1)/0.047\]

OpenStudy (anonymous):

Yeah, seems to be correct

OpenStudy (anonymous):

Excellent, Indeed \[t=\ln (\frac{3}{1.1})\frac{1}{0.047}\]

OpenStudy (anonymous):

\[t=\frac{\ln(3)-\ln(1.1)}{0.047}\]

OpenStudy (anonymous):

I have serious trouble with word problems :(

OpenStudy (anonymous):

If you don't have a calculator or you are not allowed, you'd have to convert that to base 10 \[\log_{10}\] the standard logarithm and look up the values from a log table

OpenStudy (anonymous):

Thank you for helping me

OpenStudy (anonymous):

remember this formula \[\log_{e}(x)=2.303\log_{10}(x)\]

OpenStudy (anonymous):

In general, \[\log_{b}(x)=\frac{\log_{a}(x)}{\log_{a}(b)}\] \[\frac{1}{\log_{10}(e)}=2.303\]

OpenStudy (anonymous):

Ok

OpenStudy (anonymous):

\[\log_{e}(x)=\frac{\log_{10}(x)}{\log_{10}(e)}=\frac{1}{\log_{10}(e)} \times \log_{10}(x)=2.303\log_{10}(x)\]

OpenStudy (anonymous):

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