Question

Natural log probs w/derivatives.

I'm stuck on prob. 5, this is what I did:

2lnx=ln4
lnx=ln4/2
....idk if this is right:
e^(ln4/2)=x...
the answer is 2 though so if someone could explain that one i'd appreciate it!


*postin pic

Answer 1

\[\Large 2 \ln x\quad=\quad \ln 4\]We want to start by using a rule of logarithms:\[\Large \color{royalblue}{b\cdot\log(a)\quad=\quad \log(a^b)}\]We can use this rule to deal with the 2 coefficient.

Answer 2

Gives us this, yes? :x\[\Large \ln x^2\quad=\quad \ln4\]

Answer 3

When you do the fancy exponentiation thing from there, it should clean up a lot nicer.

Answer 4

Oh but to also answer your question, yes you came up with the correct solution it just needs a lot more simplification.

\[\LARGE x\quad=\quad e^{\frac{1}{2}\ln4}\quad=\quad e^{\ln4^{1/2}}\quad=\quad 4^{1/2}\quad=\quad 2\]

Answer 5

ohh i see what you did there! Ok and then you always cancel e and ln when it's e^lnx?

Answer 6

Yes! They "undo" one another or if you like, "cancel", since the exponential and log functions are inverses of each other.

Answer 7

ok thank you for clearing that for me!

Answer 8

Here is something to be careful of though.
You `can not` "cancel" them if there is a coefficient in the way.

Example:\[\LARGE e^{2\ln x}\quad\ne\quad 2x\]You must bring the 2 into the log using log rules before you "cancel".\[\LARGE e^{2\ln x}\quad=\quad e^{\ln x^2}\quad=\quad x^2\]

Answer 9

if I may, with like the next one (6) is this right so far: \[\ln \frac{ x }{ (1-x) }=2\] then\[e ^{2}=\frac{ x }{ (x-1) }\]
\[x=e ^{2}(1-x)\] the answer is \[\frac{ e ^{2} }{ 1+e ^{2} }\] and i'm not really getting that

Answer 10

Ok your work looks great so far!
We just have a couple of weird algebra steps we can do to clean it up.

Answer 11

From this step:\[\Large x\quad=\quad e ^{2}(1-x)\]W;e can distribute the e^2 to each term in the brackets,\[\Large x\quad=\quad e^2-xe^2\]Let's get all of our x terms on one side,\[\Large x+xe^2\quad=\quad e^2\]From here we can factor an x out of each term on the left, and it isn't too bad from there.
Understand where this is going? :o

Answer 12

lol wow i feel dumb. I distributed in my head and somehow my brain told me if i do that i woudln't be able to get x alone (i was thinking x^2 so that i'd still have an x in the equ...it that makes sense haha)

Answer 13

Thank you so much!!

Answer 14

np \c:/