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We were given 1 example to do math questions. But I don't get how I got the answer.
With with Log and Ln, but in this one it's Ln.
2.6=1.4e^(0.3t)
I have to solve for t.
So you have to divide both sides by 1.4 and add Ln as well.
Ln(2.6/1.4) = Ln e^(0.3t)
Now my example in my book does not make sense. I did not make a note next to it.
What are the next examples to get to the answer of t=2.06.
If someone would kindly help me out that would be appreciated.
Thanks
Joanne
Quote:
Originally Posted by bradycat
You don't "add" log or ln to both sides: you apply it to both sides, and that only if you're sure they are positive, so:
$\displaystyle \displaystyle{\ln\left(\frac{2.6}{1.4}\right)=\ln( e^{0.3t})=0.3t\Longrightarrow t=\frac{\ln\left(\frac{2.6}{1.4}\right)}{0.3}}$ , as
$\displaystyle \ln e^x=x$ since $\displaystyle e^x\,,\ln x$ are inverse function to each other.
Tonio
The functions $\displaystyle \ln x$ and $\displaystyle e^x$ are inverse, i.e., $\displaystyle \ln(e^x)=x$ for all x and $\displaystyle e^{\ln x}=x$ for all $\displaystyle x>0$. Therefore, $\displaystyle \ln(2.6/1.4) = \ln e^{0.3t}=0.3t$, from which t can be found.
You are on the right track,
$\displaystyle 2.6 = 1.4 e^{0.3t}$
$\displaystyle \frac{2.6}{1.4}= e^{0.3t}$
$\displaystyle \ln \frac{2.6}{1.4}= 0.3t$
$\displaystyle \displaystyle\frac{\ln \frac{2.6}{1.4}}{0.3}= t$
OK I get that now. But what happens to the e on the 2nd line?? Is that the e that is applied to the left side as Ln??? Do I have that correct?
Thank you all for the help to clear this up.
Joanne
OK I called a friend of mine and the Ln is e with a base of e, so the answer is 1. So it's not needed in a way, correct?
They are the inverse of each other.
I.e $\displaystyle x+3-3 = x-3+3 = x$
$\displaystyle \displaystye \ln e^x = e^{\ln x} = x$
and also,
$\displaystyle \displaystye \log_{a} a^x = a^{\log_{a} x} = x$
