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Math Help - Some simple logarithm questions

  1. #1
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    Some simple logarithm questions

    the book simply says find the exact value of each expression. without a calculator. using properties of logarithms.( which it lists, but I can't see where they apply)
    1. log6^18-log6^3
    for this I just divided 18 by 3 getting 6. log6^6=1? is that right?
    2.log3^8 * log8^9
    for this i tried change of base. I did
    (logb^8/logb^3) * (logb^9/logb^8)
    3. e^(log(e^2)^16)
    I'm completely lost on this one. Supposedly these are all really easy to do without a calculator. i just don't know how to do them. Any help would be greatly appreciated.
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by bilbobaggins View Post
    the book simply says find the exact value of each expression. without a calculator. using properties of logarithms.( which it lists, but I can't see where they apply)
    1. log6^18-log6^3
    for this I just divided 18 by 3 getting 6. log6^6=1? is that right?
    2.log3^8 * log8^9
    for this i tried change of base. I did
    (logb^8/logb^3) * (logb^9/logb^8)
    3. e^(log(e^2)^16)
    I'm completely lost on this one. Supposedly these are all really easy to do without a calculator. i just don't know how to do them. Any help would be greatly appreciated.
    Are these logs to base ten?

    If the one problem is this \log_{6}6=y

    then by the inverse property we have 6^y=6\Longleftrightarrow{y=1}

    The logarithm of x to the base b is written logb(x) or, if the base is implicit, as log(x). So, for a number x, a base b and an exponent y,


    An
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  3. #3
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    Quote Originally Posted by VonNemo19 View Post
    Are these logs to base ten?

    If the one problem is this \log_{6}6=y

    then by the inverse property we have 6^y=6\Longleftrightarrow{y=1}
    No they're not base 10. I'm sorry, the bases are supposed to be the numbers immediately after the log. I just didn't know how to write it on a computer.
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  4. #4
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    Quote Originally Posted by bilbobaggins View Post
    the book simply says find the exact value of each expression. without a calculator. using properties of logarithms.( which it lists, but I can't see where they apply)
    1. log6^18-log6^3
    for this I just divided 18 by 3 getting 6. log6^6=1? is that right?
    2.log3^8 * log8^9
    for this i tried change of base. I did
    (logb^8/logb^3) * (logb^9/logb^8)
    3. e^(log(e^2)^16)
    I'm completely lost on this one. Supposedly these are all really easy to do without a calculator. i just don't know how to do them. Any help would be greatly appreciated.
    1. Correct.

    2. Change of base is the right way to go here but you pick a given value of b: usually base e or base 10:

    log_3{8} = \frac{ln8}{ln{3}} = \frac{3ln2}{ln3}

    log_8{9} = \frac{ln{9}}{ln{8}} = \frac{2ln{3}}{3ln2}

    \frac{3ln2}{ln{3}} \times \frac{2ln{3}}{3ln2} = 2

    3. Do you mean?

    e^{log_{e^2}16} -> the base of the log is e^2

    it's an unusual base so I'd use the change of base as above ignoring the base of the exponent for now (the e at the start). I am choosing base e because then it will cancel at the end.

    log_{e^2}16

    \frac{ln{16}}{ln{e^2}} = \frac{4ln(2)}{2ln(e)} = {2ln(2)} = ln(4)

    Bringing in our exponent we get e^{ln(4)} = 4
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Properties:
    Logarithmic identity







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  6. #6
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    Quote Originally Posted by e^(i*pi) View Post
    1. Correct.

    2. Change of base is the right way to go here but you pick a given value of b: usually base e or base 10:

    log_3{8} = \frac{ln8}{ln{3}} = \frac{3ln2}{ln3}

    log_8{9} = \frac{ln{9}}{ln{8}} = \frac{2ln{3}}{3ln2}

    \frac{3ln2}{ln{3}} \times \frac{2ln{3}}{3ln2} = 2

    3. Do you mean?

    e^{log_{e^2}16} -> the base of the log is e^2

    it's an unusual base so I'd use the change of base as above ignoring the base of the exponent for now (the e at the start). I am choosing base e because then it will cancel at the end.

    log_{e^2}16

    \frac{ln{16}}{ln{e^2}} = \frac{4ln(2)}{2ln(e)} = {2ln(2)} = ln(4)

    Bringing in our exponent we get e^{ln(4)} = 4
    So does an ln cancel out an e?

    For example in these two.

    lne^-4 and e^ln8
    By the way, thank you so much for the help.
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  7. #7
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    Quote Originally Posted by bilbobaggins View Post
    So does an ln cancel out an e?

    For example in these two.

    lne^-4 and e^ln8
    By the way, thank you so much for the help.
    Yes, e and ln are inverse operators, like addition and subtraction etc...

    ln(x) is just a different way of saying log_e(x)

    ln(e^x) = e^{ln(x)} = x

    So ln(e^{-4}) = -4 and e^{ln(8)} = 8
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  8. #8
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    I have another one that's really confusing.

    21log3^cuberootx +log3^(9x^2)-log3^9

    I'm supposed to get it to a single logarithm. I get it down to log3^(x^20). Is that right? Or is it log3^(9x^20/9)?
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  9. #9
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    Quote Originally Posted by bilbobaggins View Post
    I have another one that's really confusing.

    21log3^cuberootx +log3^(9x^2)-log3^9

    I'm supposed to get it to a single logarithm. I get it down to log3^(x^20). Is that right? Or is it log3^(9x^20/9)?
    21log_3(x^{\frac{1}{3}}) = log_3(x^{\frac{21}{3}}) + log_3(9x^2) - log_3(9)

    = log_3(7\times 9x^2 \div 9) = log_3(7x^2)

    That will need to be simplified to give a final answer
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  10. #10
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    Quote Originally Posted by e^(i*pi) View Post
    21log_3(x^{\frac{1}{3}}) = log_3(x^{\frac{21}{3}}) + log_3(9x^2) - log_3(9)

    = log_3(7\times 9x^2 \div 9) = log_3(7x^2)

    That will need to be simplified to give a final answer
    I'm a little lost on the second line. So the x^21/3 becomes just seven? Because of 21/3?
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  11. #11
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    Quote Originally Posted by bilbobaggins View Post
    I'm a little lost on the second line. So the x^21/3 becomes just seven? Because of 21/3?
    Damn, that's a typo, It should be x^7 rather than 7. The exponent of 7 does come from the 21/3

    <br />
= log_3(x^7 \times 9x^2 \div 9) = log_3(x^9)<br />

    You multiply because of the rule that says log(a)+log(b) = log(ab). The 9/9 will cancel inside the log term
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  12. #12
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    Quote Originally Posted by e^(i*pi) View Post
    Damn, that's a typo, It should be x^7 rather than 7. The exponent of 7 does come from the 21/3

    <br />
= log_3(x^7 \times 9x^2 \div 9) = log_3(x^9)<br />

    You multiply because of the rule that says log(a)+log(b) = log(ab). The 9/9 will cancel inside the log term
    Alright, thank you very much for the help. It's a lot clearer now. I wish you were my math instructor!
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