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
do you mean \[P(t)=1.1e^{0.047t}\] or \[P(t)=1.1\times e \times 0.047t\]
the first one sorry
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?
Not really :(
I would divide 3 by 11
Have you not read about logarithms?
I have but I'm having hard time understanding it
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
ok
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
Ok, Thank you
\[\log_{e}(e)=1\] and \[\log_{e}(e^x)=xlog_{e}(e)=x\]
the answer I got was 21.35...
let me check
\[T = \ln (3/1.1)/0.047\]
Yeah, seems to be correct
Excellent, Indeed \[t=\ln (\frac{3}{1.1})\frac{1}{0.047}\]
\[t=\frac{\ln(3)-\ln(1.1)}{0.047}\]
I have serious trouble with word problems :(
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
Thank you for helping me
remember this formula \[\log_{e}(x)=2.303\log_{10}(x)\]
In general, \[\log_{b}(x)=\frac{\log_{a}(x)}{\log_{a}(b)}\] \[\frac{1}{\log_{10}(e)}=2.303\]
Ok
\[\log_{e}(x)=\frac{\log_{10}(x)}{\log_{10}(e)}=\frac{1}{\log_{10}(e)} \times \log_{10}(x)=2.303\log_{10}(x)\]
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