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Equation: $(25+e^{4t})(2e^{2t}) - \frac{3}{2}(4e^{4t}) = 0$

I'm having some problems solving for t. These are the steps I attempt to take:

$50e^{2t} + 2e^{6t} - 6e^{4t} = 0$

$50 + 2e^{4t} - 6e^{2t} = 0$

$2e^{4t} - 6e^{2t} = -50$

$e^{4t} - 3e^{2t} = -25$

At this step, I was planning to ln:

$\ln(e^{4t}) - \ln(3e^{2t}) = \ln(-25)$

BUT, you cannot ln an negative number, so I have run into a problem and am not sure how to go about solving it. Any tips or advice pointing me in the right direction is appreciated!

user
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Amai
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    The logarithm of a difference is not the difference of logarithms. You need another strategy. Hint: let $z = e^{2t}$. Find $z$ and then use logarithms. – Ethan Bolker Sep 29 '19 at 00:10
  • another way to solve is by thinking how to use functions property to solve that, define the function according to the equation, and check its derivative – friedvir Sep 29 '19 at 00:14

3 Answers3

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Firstly, you can take the log of a negative number - but the biggest problem with your method is that you are performing $$ \ln(a + b) \neq \ln a + \ln b $$ This is not a linear function.

So to solve the above you need to make a sub $$ y = \mathrm{e}^{2t} $$ then solve $$ y^2-3y = 50 $$ then find $t$ from $y$

Chinny84
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  • When I try to solve it through substitution, it seems that I cannot end up with a real number. Is it possible for you to confirm if this is the case because it is quite confusing for me? The number was eventually supposed to be plugged in to find the maximum curvature for a space curve, so I don't believe I can use an unreal number for that. – Amai Sep 29 '19 at 00:37
  • as it stands you won't have real solutions as another poster showed. I think you will have to show how you got to the point above and try to see if that was valid. If that holds up then you will have to figure out how the solution is valid. – Chinny84 Sep 29 '19 at 00:43
  • Okay. Uh...do you happen to know if I should create a new post to show my base work or just edit my current one? Q A Q – Amai Sep 29 '19 at 00:44
  • I think it is best to create a new post - you could link this post as the motivation for this question. – Chinny84 Sep 29 '19 at 00:49
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Let $x=e^{2t}$ then

$$e^{4t} - 3e^{2t} = -25 \iff x^2-3x+25=0$$

Note that in you trial, the correct step should be

$$e^{4t} - 3e^{2t} = -25 \iff 3e^{2t}-e^{4t} =25 \iff \ln{\left(3e^{2t}-e^{4t}\right)}=\ln 25$$

which is in any case not conclusive.

user
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To begin with, notice that

\begin{align*} 2(25+e^{4t})e^{2t} - 6e^{4t} = 0 \Longleftrightarrow (25 + e^{4t}) - 3e^{2t} = 0 \Longleftrightarrow e^{4t} - 3e^{2t} + 25 = 0 \end{align*}

which has no real solutions, since $\Delta < 0 $.

user0102
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