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Math Help - Simplyfying Logarithms?

  1. #1
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    Simplyfying Logarithms?

    First we have: 8 Ln(e^a) + 9 ln(B^e)

    8a Ln(e) + 9 ln(B^e)

    8a + 9ln(B^e)

    Anybody know what to do from here?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Gfrant7 View Post
    First we have: 8 ln(eA) + 9 ln(Be)

    8a Ln(e) + 9 ln(B^e)

    8a + 9ln(B^e)

    Anybody know what to do from here?
    wait, i'm confused. it seems you are working with powers in the second line, yet in the first line you only have products. which is it?

    do you mean 8 \ln (e^A) + 9 \ln (B^e) ?

    if so, you should indicate powers by using "^" just as you did in the second line
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  3. #3
    Moo
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    Hello,

    The only thing you can do is ln(b^e)=e*ln(b).

    What was it for ?
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    Sorry Jhevon, I didn't have it written correctly in my initial post. That is what I meant though.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Gfrant7 View Post
    Sorry Jhevon, I didn't have it written correctly in my initial post. That is what I meant though.
    well, in that case, follow Moo's post. you can't really to anything more than changing \ln (B^e) to e \ln B

    if you were told the values of ln(B) at some point, you can plug it in, if not, just leave it
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  6. #6
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    In that case we would get

    8ln(e^a) + 9ln(b^e)

    which would simplify to

    8a + 9eln(b) which webassign is telling me is the incorrect answer. damn this problem lol
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  7. #7
    Moo
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    Oh !

    I think I can see what you're asked.

    use these rules a \ln b=\ln(b^a) and \ln(a)+\ln(b)=\ln(ab) and (a^b)^c=a^{bc}

    --> 8 \ln(e^a)=\ln((e^a)^8)=\ln(e^{8a})

    --> 9 \ln(b^e)=\ln(\dots)

    -----> 8 \ln(e^a)+9 \ln(b^e)=\ln(\dots)+\ln(\dots)=\ln(\dots \dots)
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Moo View Post
    Oh !

    I think I can see what you're asked.

    use these rules a \ln b=\ln(b^a) and \ln(a)+\ln(b)=\ln(ab) and (a^b)^c=a^{bc}

    --> 8 \ln(e^a)=\ln((e^a)^8)=\ln(e^{8a})

    --> 9 \ln(b^e)=\ln(\dots)

    -----> 8 \ln(e^a)+9 \ln(b^e)=\ln(\dots)+\ln(\dots)=\ln(\dots \dots)
    Oh! haha, ok! that's makes sense

    you should have told us you were trying to simplify into a single logarithm, Gfrant7
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  9. #9
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    Thanks! would it be

    ln(e^8a) + ln(b^9e)

    = ln((e^8a)*(b^9e))
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Gfrant7 View Post
    Thanks! would it be

    ln(e^8a) + ln(b^9e)

    = ln((e^8a)*(b^9e))
    yes, but make sure your syntax is correct if you are typing it. you might need the power in brackets, otherwise it might interpret b^9e as b^9e when you mean b^{9e}. and so on...
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