Solving $7 = 8 - 2e^{-3k}$

Question

So, this is an assignment my friend and I have for our homework: $7 = 8 - 2e^{-3k}$

And the solution should be: $\frac{1}{3\ln(2)}$

But, I have no idea how they got there. I tried doing:

$$\begin{align*} 7 &= 8 - 2e^{-3k}\\ 1 &= 2e^{-3k}\\ \frac{1}{2} &= e^{-3k} \end{align*}$$

And now I don't know what to do next. Help!

PS: Neither of us was in school when they were learning this lesson. So, some description along the way will be greatly welcomed!

From your last equation: remember that $\ln,e^u=u$... Also, $\ln\frac1{s}=-\ln,s$. — J. M. ain't a mathematician, Sep 15 '11 at 19:14
Take reciprocals: $e^{3k} = 2$. Take logs: $3k = \ln 2$. Solve for $k$. — Srivatsan, Sep 15 '11 at 19:16
(You should probably review what a logarithm is and its basic properties.) — anon, Sep 15 '11 at 19:21
But this does not lead to the specified conclusion. Perhaps there was an error copying the assignment and/or solution? — hmakholm left over Monica, Sep 15 '11 at 19:22
Yes, thought if you are sure this is the correct question, then you copied the answer wrong. — Leonardo Fontoura, Sep 15 '11 at 19:26

score 1 · Accepted Answer · answered Sep 15 '11 at 19:56

1

$$ \begin{align} \frac12 & = e^{-3k} \\ \\ 2 & = e^{3k} \\ \\ \log_e 2 & = 3k \\ \\ \frac{\log_e 2}{3} & = k \end{align} $$

answered Sep 15 '11 at 19:56

Michael Hardy

1

Oh, we were getting that answer, too. I guess she copied down the wrong solution or something. Haha. Thanks! – omninonsense Sep 15 '11 at 20:23

score 0 · Answer 2 · answered Sep 15 '11 at 19:24

The logarithm of a positive real number $x$ with respect to the base $a \in (0,\infty)\setminus \{1\}$ is the unique real number $\log_a x$ with the property that $a^{\log_a x}=x$.

In your case you have $\frac{1}{2}=e^{-3k}$ which is equivalent to $e^{3k}=2$. $\ln$ is the notation for $\log_e$ so by the definition in the first paragraph $3k=\ln e^{3k}=\ln 2$.

Solving $7 = 8 - 2e^{-3k}$

2 Answers2