How to solve for t in the equation mentioned below?

Question

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!

Answer 1

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 $y$

Answer 2

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 $.

Answer 3

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.