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anonymous

  • 5 years ago

can someone help w/ logs

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  1. anonymous
    • 5 years ago
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    example please

  2. anonymous
    • 5 years ago
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    what would you like to know? the main thing to remember is that logs are actually exponents (until you get to fancy math) so all the laws of exponents apply.

  3. anonymous
    • 5 years ago
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    \[3^ (x+1) = 6^x\]

  4. anonymous
    • 5 years ago
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    wait thats wrong what i posted

  5. anonymous
    • 5 years ago
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    \[3^{x+1}=6^x\]

  6. anonymous
    • 5 years ago
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    its 3 ^ (x+1) = 6^x

  7. anonymous
    • 5 years ago
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    yes that

  8. anonymous
    • 5 years ago
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    has nothing to do with logs.

  9. anonymous
    • 5 years ago
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    oops i lied excuse me

  10. anonymous
    • 5 years ago
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    yes it does you have to multiply each side by a log...huh?

  11. anonymous
    • 5 years ago
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    no you do not 'multiply' by the log you 'take the log'

  12. anonymous
    • 5 years ago
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    you can choose any base you like, but I will use \[log_e\] which is written \[ln\]

  13. anonymous
    • 5 years ago
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    thats what i mean...im so confused

  14. anonymous
    • 5 years ago
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    take the log of both sides to get \[ln(3^{x+1})=ln(6^x)\]

  15. anonymous
    • 5 years ago
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    then what? log and what base? dont u mean log3 (log base 3 i mean)

  16. anonymous
    • 5 years ago
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    then the exponents come out as multipliers to give \[(x+1)ln(3)=xln(6)\] we solve this equation for x, remembering that \[ln(3)\] and \[ln(6)\] are just constants.

  17. anonymous
    • 5 years ago
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    you "take" ln and use a ln properties... (x+1)ln3=xln6 (x+1)/x = ln6/ln3 1+ 1/x=ln6/ln3 1/x=ln6/ln3 -1 x=ln3/(ln6-ln3) you can use log with base of 3 as well...

  18. anonymous
    • 5 years ago
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    let me start slowly. You have two different bases here, 3 and 6. whatever you do to one side you must do the same to the other. you cannot take the log base 3 of one side and the log base 6 of the other.

  19. anonymous
    • 5 years ago
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    ok what do i do after (x+1) log (3) = x log 6

  20. anonymous
    • 5 years ago
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    both sides... 6=3*2, so you can present log3 (3*2)= 1+log3(2) (log3 - log with base of 3)

  21. anonymous
    • 5 years ago
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    ahhh too many people are helping me im so confused

  22. anonymous
    • 5 years ago
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    the same as in ln... you'll have ln or log in answer

  23. anonymous
    • 5 years ago
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    let me see if i can unconfuse you. how do you solve any equation with the variable in the exponent? for example how would you solve \[2^x=1000\]?

  24. anonymous
    • 5 years ago
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    you cannot use 'algebra' because the variable is in the sky. it is in the exponent not on the ground floor, so you cannot add or subtract, multiply or divide to solve for x.

  25. anonymous
    • 5 years ago
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    you solve it by taking the log of both sides, because one of the facts of logs is that \[log_b(a^x)=xlog_b(a)\] and now the variable is on the ground floor. so can solve using algebra

  26. anonymous
    • 5 years ago
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    so if i want to solve \[2^x=1000\] for x i take the log of both sides and get \[log(2^x)=log(1000\] \[xlog(2)=log(1000)\] \[x=\frac{log(1000)}{log(2)}\]

  27. anonymous
    • 5 years ago
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    now you have \[3^{x+1}=6^x\] tale the log of both sides to get \[log(3^{x+1})=log(6^x)\] \[(x+1)log(3)=xlog(6)\] multiply out to get \[x\log(3)+log(3)=xlog(6)\] \[log(3)=xlog(6)-xlog(3)\] \[log(3)=x(log(6)-log(3))\] \[\frac{log(3)}{log(6)-log(3)}=x\]

  28. anonymous
    • 5 years ago
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    which base is the log? it doesn't matter. but if you want a number out of this you have to use a base that you can find on your calculator. you can use log base ten which is just written as log or you can use log base e which is written as ln.

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