## animalsavior94 Group Title why isnt anyone helping me?! 3 years ago 3 years ago

1. malevolence19 Group Title

2. animalsavior94 Group Title

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

3. malevolence19 Group Title

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?

4. malevolence19 Group Title

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

5. animalsavior94 Group Title

im so screwed for the exam:(

6. malevolence19 Group Title

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

7. animalsavior94 Group Title

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

8. malevolence19 Group Title

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

9. animalsavior94 Group Title

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

10. malevolence19 Group Title

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 :)

11. animalsavior94 Group Title

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

12. malevolence19 Group Title

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

13. animalsavior94 Group Title

will do so:]

14. animalsavior94 Group Title

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

15. animalsavior94 Group Title

condense it to a log of a single quantity

16. malevolence19 Group Title

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

17. malevolence19 Group Title

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: $\ln(x^{1/2}(x+3))-\ln(x^2+1)=\ln(\frac{x^{1/2}(x+3)}{x^2+1})$

18. malevolence19 Group Title

Make sense?