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animalsavior94

  • 3 years ago

why isnt anyone helping me?!

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  1. malevolence19
    • 3 years ago
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    Your question?

  2. animalsavior94
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    Forgot one: \[a^{\log_a(x)}=x\] (same as one of the above but in reverse)

  5. animalsavior94
    • 3 years ago
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    im so screwed for the exam:(

  6. malevolence19
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    Well find some questions you don't understand and I'll try to explain if you want :)

  9. animalsavior94
    • 3 years ago
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    aww you would do that to help me? thanks sooo much! i appreciate it greatly!!:D

  10. malevolence19
    • 3 years ago
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    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
    • 3 years ago
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    that sounds pretty good! thank you soooo much once again!!

  12. malevolence19
    • 3 years ago
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    No problem :) Just post whatever you get stuck on.

  13. animalsavior94
    • 3 years ago
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    will do so:]

  14. animalsavior94
    • 3 years ago
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    \[1/2\ln x+\ln (x+3)-\ln (x ^{2}+1)\]

  15. animalsavior94
    • 3 years ago
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    condense it to a log of a single quantity

  16. malevolence19
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    Make sense?

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