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Math Help - Please help me solve the ff Log Problems.

  1. #1
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    Unhappy Please help me solve the ff Log Problems.

    Without the use of a calculator.

    1. e^ln0.1
    2. log(base)2 3. log(base)3 4. log(base)4 8
    3. e^log(base)e^2^9
    4. ln(y+4)=5x+lnC
    5. log(base)5(x+6)+log(base)5 (x+1)=1
    Determine the domain of the ff:
    f(x)=log(base)2 (xsquare-3x+2).

    When i say log(base)2, I mean the no that goes underneath the log.
    Thanks a lot!
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  2. #2
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    Quote Originally Posted by different92 View Post
    Without the use of a calculator.

    1. e^ln0.1
    2. log(base)2 3. log(base)3 4. log(base)4 8
    3. e^log(base)e^2^9
    4. ln(y+4)=5x+lnC
    5. log(base)5(x+6)+log(base)5 (x+1)=1
    Determine the domain of the ff:
    f(x)=log(base)2 (xsquare-3x+2).

    When i say log(base)2, I mean the no that goes underneath the log.
    Thanks a lot!
    Some hints

    Inverse functions: e^{\ln(a)} = \ln(e^a) = a

    ^ in general \log_a(a^k) = a^{\log_a(k)} = k

    Change of base rule: \log_b(a) = \frac{\log_c(a)}{\log_c(b)}

    Addition Rule: \log_b(a) + \log_b(c) = \log_b(ac)

    Subtraction Rule: \log_b(a) - \log_b(c) = \log_b\left(\frac{a}{c}\right)

    Power Law: \log_a(b^k) = k\,\log_a(b)
    ---------------------------

    For example number 2 (I apologise if I have misread the question but the syntax is highly confusing)

    From the change of base rule applied to each term (I have used base e but it works with any positive base that is not 1): \log_2(3) \cdot \log_3(4) \cdot \log_4(8) = \frac{\ln(3)}{\ln(2)} \cdot \frac{\ln(4)}{\ln(3)} \cdot \frac{\ln(8)}{\ln(4)}

    since 8 = 2^3 and 4 = 2^2 we can simplify using the power law


    \frac{\ln(3)}{\ln(2)} \cdot \frac{2\ln(2)}{\ln(3)} \cdot \frac{3\ln(2)}{2\ln(2)}

    Cancel out terms as you would with any ordinary fraction to give an answer of 3
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