Post by: K.Smithy on December 24, 2019, 04:17:27 pm
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)(e2x - 4) = 0 for x
I've used null factor, so what I've done looks like this...
e-x - 1 = 0 OR e2x - 4 = 0
e-x = 1 e2x = 4
logee-x = ln(1) logee2x = 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:
e2x = 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
Post by: MB_ on December 24, 2019, 04:22:26 pm
In this case \begin{align*}
\ln(4)=\ln(2^2)\\
=2\ln(2)
\end{align*}
Post by: Sine on December 24, 2019, 04:23:27 pm
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)
Post by: K.Smithy on December 24, 2019, 04:29:10 pm
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 ;D Thank you so much MB_ and Sine! :D