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Math Help - logarithmic functions

  1. #1
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    Post logarithmic functions

    We have just started this chapter and I am confused on the more complicated problems. The book has omitted the way this is worked out. Could you please help explain the steps to me?

    e^3In2-2In5 = 8/25
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  2. #2
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by confusedagain View Post
    We have just started this chapter and I am confused on the more complicated problems. The book has omitted the way this is worked out. Could you please help explain the steps to me?

    e^3In2-2In5 = 8/25
    I assume you need to show that the two sides are equal.

    e^(3ln2-2ln5) = e^(ln2^3 - ln5^2) = e^(ln8 - ln25) = e^(ln(8/25) = 8/25

    If you need further explanation of certain steps, say so and I'll be happy to help.
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    I assume you need to show that the two sides are equal.

    e^(3ln2-2ln5) = e^(ln2^3 - ln5^2) = e^(ln8 - ln25) = e^(ln(8/25) = 8/25

    If you need further explanation of certain steps, say so and I'll be happy to help.
    I'll go ahead and explain the steps:

    First, we have:
    e^(3ln2 - 2ln5)

    In both terms we have a number being multiplied to the log. We can take each number and make it the exponent of the term inside the log. In other words,
    3ln2 = ln2^3 = ln8
    2ln5 = ln5^2 = ln25

    So we get:
    e^(3ln2 - 2ln5) = e^(ln8 - ln25)

    Since we have a subtraction of logs, we can combine this into a common log by dividing the terms of the logs. In other words,
    ln8 - ln25 = ln(8/25)

    So we get:
    e^(3ln2 - 2ln5) = e^(ln8 - ln25) = e^(ln(8/25))

    Now that we have "e" raised to the ln of some term, we can "cancle" the e and ln since they are considered "inverse opperations." In other words,
    e^(lnx) = x
    ln(e^x) = x

    So we get:
    e^(3ln2 - 2ln5) = e^(ln8 - ln25) = e^(ln(8/25)) = 8/25
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  4. #4
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    Post I think I understand but...

    I think I understand these types of problems now.

    So if the problem is e^2ln3 = 3ln2=ln3^2 = 9

    What about a problem such as this:
    ln e(^3square root e/e^1/3)?
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by confusedagain View Post
    I think I understand these types of problems now.

    So if the problem is e^2ln3 = 3ln2=ln3^2 = 9

    What about a problem such as this:
    ln e(^3square root e/e^1/3)?

    First of all, write out your problems clearly. e^(2ln3) is NOT equal to 3ln2 or ln3^2


    you should have written e^(2ln3) = e^(ln(3^2)) = e^(ln9) = 9


    keep brackets around your powers if they are expressions, we need to know where they start or stop.




    Now we have lne^(3sqrt(e)/e^(1/3))


    Now there is a law of logarithms that says: log(x^n) = nlogx


    so lne^(3sqrt(e)/e^(1/3)) = (3sqrt(e)/e^(1/3))*lne


    there is also a law that says log[a]a = 1, i put the base in [] so i just wrote log to the base a of a = 1
    so if the base is the same as the number being logged, the answer is one, i hope you know why (if not ask). Now ln = log[e], so lne = log[e]e = 1 (we don't write log[e]e, we just go straight to 1)


    so we have lne^(3sqrt(e)/e^(1/3)) = (3sqrt(e)/e^(1/3))*lne = 3sqrt(e)/e^(1/3)


    now sqrt(e) = e^(1/2). A law of exponents says that if we divide a base into itself, we subtract the power of the bottom from the one at the top.


    So 3sqrt(e)/e^(1/3) = [3e^(1/2)]/e^(1/3) = 3* e^(1/2) / e^(1/3) = 3 * e^(1/2 – 1/3) = 3e^(1/6) and that's your answer
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