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Math Help - logarithm

  1. #1
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    Exclamation logarithm

    I need help!

    how to solve

    log10x^7.. expanding with the rules..

    9e^5x=90

    and 2^6-x=10

    I would greatly appreciate the help! I know you guys were so urgent to help last time I needed it!

    TAHNKSS!!
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  2. #2
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    Quote Originally Posted by Mmarie View Post
    I need help!

    how to solve

    log10x^7.. expanding with the rules..

    9e^5x=90

    and 2^6-x=10

    I would greatly appreciate the help! I know you guys were so urgent to help last time I needed it!

    TAHNKSS!!
    Please clarify:

    do you mean \log_{10}x^7

    and

    2^{6 - x} = 10?

    For the other question:

    9e^{5x} = 90

    \Rightarrow e^{5x} = 10

    \Rightarrow \ln e^{5x} = \ln 10

    \Rightarrow 5x \ln e = \ln 10 .................since log_a (x^n) = n \log_a x

    \Rightarrow 5x = \ln 10 ....................since \log_a a = 1 so in particular, \ln e = 1

    \Rightarrow x = \frac {\ln 10}5
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  3. #3
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    its log 10.. not subscript 10...
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    Quote Originally Posted by Mmarie View Post
    its log 10.. not subscript 10...
    \log 10x^7 = \log 10 + \log x^7 ..............since \log_a xy = \log_a x + \log_a y

    = \log 10 + 7 \log x .................since \log_a (x^n) = n \log_a x

    = 1 + 7 \log x .......................if the log is the base 10 logarithm. since \log_a a = 1




    2^{6 - x} = 10

    \Rightarrow \ln 2^{6 - x} = \ln 10

    \Rightarrow (6 - x) \ln 2 = \ln 10

    \Rightarrow 6 - x = \frac {\ln 10}{\ln 2}

    \Rightarrow x = 6 - \frac {\ln 10}{\ln 2}
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  5. #5
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    I also don't understand condensing and expanding.. at all.. my teacher is not good, at all. he doesn't explain how to work it. he puts up the slides and doesn't explain it.
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    Quote Originally Posted by Mmarie View Post
    I also don't understand condensing and expanding.. at all.. my teacher is not good, at all. he doesn't explain how to work it. he puts up the slides and doesn't explain it.
    i provided you with the rules when i answered the question so that you know how i am expanding it. at this point, since you have a test soon. it is best to just memorize these and apply them.
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  7. #7
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    when do you know when to use ln and log?
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    Quote Originally Posted by Mmarie View Post
    when do you know when to use ln and log?
    if you are working with base e, use ln. if you are working with base 10, use log. Otherwise it does not matter. But it is customary to use ln in those cases, at least in calculus. precalculus and below perhaps use log (to the base 10) as the standard, i don't remember.
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  9. #9
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    Hello,

    Quote Originally Posted by Mmarie View Post
    I also don't understand condensing and expanding.. at all.. my teacher is not good, at all. he doesn't explain how to work it. he puts up the slides and doesn't explain it.
    And in general, you have few rules :

    \ln(ab)=\ln(a)+\ln(b). Plus, \ln \frac ab=\ln(a)-\ln(b)

    \ln(a^b)=b \ln(a)

    \ln(e)=1

    \log_{10}(10)=1
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  10. #10
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    okay, so how do you condense..

    log5 x-2log5 y

    with 5 being subscript
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  11. #11
    Moo
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    Know this formula :

    \log_{n}(x)=\frac{\ln(x)}{\ln(n)}

    And it may be all done.
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  12. #12
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    One more question, on this section.

    I just had an epiphany, ha! I understand all except for how to condense this one.. can you provide a step by step for this one..

    log7 (x^2-9)- log7 (x-3)
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  13. #13
    Moo
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    Nope I won't

    I'd just tell you that x^2-9=(x-3)(x+3)
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  14. #14
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    Okay, I got that one! Thank you..

    Now... haha sorry..

    log3 (x+1) + log3 (x+3)=1

    the answer is there is no solution, 0, because you can't have a take a negative log.. i tried to solve it and i combined them and did log3 (x^2+4x+3)=1 but i have no idea where to go from there
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  15. #15
    Moo
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    Well, 0 is a solution
    x+1>0 and x+3>0
    So x has to be >-1

    \log_3 (x^2+4x+3)=1=\log_3(3)

    This means that x^2+4x+3=3

    x^2+4x=0 \Longleftrightarrow x(x+4)=0 \dots
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