Indicial Equations with e

[K.Smithy]

Hi everyone

I was just revising content from the end of the year and I came across one question that I'm a bit confused about.
I have to solve (e^-x - 1)(e^2x - 4) = 0 for x

I've used null factor, so what I've done looks like this...
e^-x - 1 = 0     OR      e^2x - 4 = 0
e^-x = 1                      e^2x = 4
log_ee^-x = ln(1)           log_ee^2x = ln(4)
-x = 0                       2x = 1.386
x = 0                         x = 0.693

Therefore, x = 0 or 0.693...

What the textbook has done instead is:
e^2x = 4
2x = ln(4)
This makes sense so far... but then it goes...
2x = 2ln(2)
x = ln(2)

How did they get 2ln(2)?
I've plugged it into my calculator and ln(2) = 0.693.. so my answer is correct, I just don't understand the textbooks working out....

Thanks
- Katelyn

[MB_]

They've used the log law \[\ln(a^b)=b\ln(a)\]

In this case \begin{align*}
\ln(4)=\ln(2^2)\\
=2\ln(2)
\end{align*}

[Sine]

Hi Katelyn,

Like you said your answer is definitely correct. The textbook just seems to have used a power log law.

That law is:



So in your case

ln(4) = ln(2²) = 2ln(2)

[K.Smithy]

Hi Katelyn,

Like you said your answer is definitely correct. The textbook just seems to have used a power log law.

That law is:



So in your case

ln(4) = ln(2²) = 2ln(2)

They've used the log law \[\ln(a^b)=b\ln(a)\]

In this case \begin{align*}
\ln(4)=\ln(2^2)\\
=2\ln(2)
\end{align*}

Oh lol, good old log laws... Grade 10 mathematics, yet I still manage to forget them Thank you so much MB_ and Sine!