Question

why isnt anyone helping me?!

Answer 1

Your question?

Answer 2

could you please explain the different laws or stuff you could do in logs? because i am so confused about the entire chapter logs and i got a final exam tomorrow so im stressing out

Answer 3

Okay. :)
A logarithm is the inverse of an exponential in the same sense the square root (almost) is the inverse of squaring something.
So if you have something like:
log(7x)=2
You can raise both sides by 10 so:
10^(log(7x))=10^2
Giving:
7x=100
However, the same works in reverse.
Given 10^7x=10^2
You can take the log
log(10^7x)=log(10^2)
7x=2
One nice thing about logarithms is the "base" in which if you have 2^x you can take log base TWO to simplify it to x.
There are several properties of logs.
\[\log(ab)=\log(a)+\log(b)\]
\[\log(\frac{a}{b})=\log(a)-\log(b)\]
\[\log_a(x)=\frac{\log_b(x)}{\log_b(a)}\](change of base)
\[\log_a(a^x)=x\]
NOTE: If I have log it is assumed base 10. Ln is assumed log base e (e=2.718...)
Also, a more obscure manipulation is:
\[\log_a(x)=-\log_\frac{1}{a}(x)\]
You can rewrite logs bringing powers down as coefficients.
\[\log_a(x^n)=nlog_a(x)\]
NOTE: if you have:\[\log_a^n(x)\neq nlog_a(x)\] Make sure you realize the difference.
Any questions?

Answer 4

Forgot one:
\[a^{\log_a(x)}=x\]
(same as one of the above but in reverse)

Answer 5

im so screwed for the exam:(

Answer 6

Well I can help you work through examples if you want. That's about all I know about logs except stuff involving calculus and stuff xP

Answer 7

Well the exam im taking tomorrow bright and early at 9am tomorrow is pre calc. I never understood logs ever since algebra 2 lol its really confusing especially the change base and stuff aka exponential form

Answer 8

Well find some questions you don't understand and I'll try to explain if you want :)

Answer 9

aww you would do that to help me? thanks sooo much! i appreciate it greatly!!:D

Answer 10

No problem :) I should be on for about an hr and a half. Then I'll be off for about 20 then I'll be back for another hour or so :)

Answer 11

that sounds pretty good! thank you soooo much once again!!

Answer 12

No problem :) Just post whatever you get stuck on.

Answer 13

will do so:]

Answer 14

\[1/2\ln x+\ln (x+3)-\ln (x ^{2}+1)\]

Answer 15

condense it to a log of a single quantity

Answer 16

I'm assuming you want to combine it.
Using properties you can see that:
\[\frac{1}{2}\ln(x)=\ln(x^{\frac{1}{2}})\]

Answer 17

From here you have an addition of two logs, then a subtraction of another. This means the two additions will be in the numerator of the new combined log, and the subtracted one will be in the denominator. So:
\[\ln(x^{1/2})+\ln(x+3)=\ln(x^{1/2}(x+3))\]
Then from here you have:
\[\ln(x^{1/2}(x+3))-\ln(x^2+1)=\ln(\frac{x^{1/2}(x+3)}{x^2+1})\]

Answer 18

Make sense?