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Math Help - logarithms help.

  1. #1
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    logarithms help.

    okay, here's my problem.

    ln 4 + ln x = 3

    i am enrolled in an independent study course and i am having a lot of trouble with this problem. i am completely unsure as of how to go about solving this problem. there are other problems i need help with as well. but maybe help with this one will help me figure out the others.

    any help offered would be appreciated!
    thank you. =]
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beli3ve View Post
    okay, here's my problem.

    ln 4 + ln x = 3

    i am enrolled in an independent study course and i am having a lot of trouble with this problem. i am completely unsure as of how to go about solving this problem. there are other problems i need help with as well. but maybe help with this one will help me figure out the others.

    any help offered would be appreciated!
    thank you. =]
    \ln 4 + \ln x = 3

    \Rightarrow \ln 4x = 3 ..........since \log_a xy = \log_a x + \log_a y

    \Rightarrow 4x = e^3 ..........since if \log_a b = c then a^c = b. Note that \ln \equiv log_e

    \Rightarrow x = \frac {e^3}{4}
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  3. #3
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    Quote Originally Posted by Jhevon View Post
    \ln 4 + \ln x = 3

    \Rightarrow \ln 4x = 3 ..........since \log_a xy = \log_a x + \log_a y

    \Rightarrow 4x = e^3 ..........since if \log_a b = c then a^c = b. Note that \ln \equiv log_e

    \Rightarrow x = \frac {e^3}{4}
    I am having a hard time understand
    \Rightarrow x = \frac {e^3}{4}.
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  4. #4
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    Wait, I am getting now what ln is equivalent to, rather than ln = log, I believe that's where my mistake was. But now at the same time, it was only a 3.

    How/why did it change to e^3?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beli3ve View Post
    I am having a hard time understand
    \Rightarrow x = \frac {e^3}{4}.
    i had 4x = e^3 so i just divided both sides by 4, that's all
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by beli3ve View Post
    Wait, I am getting now what ln is equivalent to, rather than ln = log, I believe that's where my mistake was. But now at the same time, it was only a 3.

    How/why did it change to e^3?
    as i told you, there is a law of logarithms that says:

    if \log_a b = c then a^c = b

    we had \ln 4x = 3 we want to find x, so i applied the above law. since we can think of ln as log to the base e, my next line holds true


    so it is like we had \log_e (4x) = 3 so e^3 = 4x
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  7. #7
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    I wasn't making the connection of that last law.
    Thank you very much. =]

    That has made my next couple problems much easier.
    You're a blessing.
    Have a great rest of the day. =]
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