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Thread: Natural and common logarithms inequalities?

  1. #1
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    Natural and common logarithms inequalities?

    I have two problems one is a common logarithm and the other is a natural logarithm.

    1. Use natural logarithms to solve each inequality
    x^(2/3) greater than to 27.6

    (2/3) ln x greater than ln 27.6
    And I am stuck right here because if I put ln(x) in my calculator
    I get 2.3025 but If proceed with this I get the wrong answer.

    2. This next problem I must solve using common logarithms

    log x^(6) greater than 1 I am not sure how to start this one
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  2. #2
    Super Member Quacky's Avatar
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    Quote Originally Posted by homeylova223 View Post
    I have two problems one is a common logarithm and the other is a natural logarithm.

    1. Use natural logarithms to solve each inequality
    x^(2/3) greater than to 27.6

    (2/3) ln x greater than ln 27.6
    And I am stuck right here because if I put ln(x) in my calculator
    I get 2.3025 but If proceed with this I get the wrong answer.

    2. This next problem I must solve using common logarithms

    log x^(6) greater than 1 I am not sure how to start this one
    $\displaystyle x^{\frac{2}{3}}>27.6$
    Take logs of both sides:

    $\displaystyle \displaystyle Log(x^{\frac{2}{3}})>Log(27.6)$

    $\displaystyle \frac{2}{3}Log(x)>Log(27.6)$

    $\displaystyle Log(x)>\frac{3}{2}Log(27.6)$

    $\displaystyle Log(x)>Log(27.6^{\frac{3}{2}})$

    $\displaystyle x>27.6^{\frac{3}{2}}$

    Try b) yourself - the method is similar!
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  3. #3
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    For my second problem would I do this

    log x^(6) greater than 1
    6 log x greater than log 1 (Then how would I proceed?)
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  4. #4
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    Quote Originally Posted by homeylova223 View Post
    For my second problem would I do this

    log x^(6) greater than 1
    6 log x greater than log 1 (Then how would I proceed?)

    The original version of the question (in the OP) states that $\displaystyle \log(x^6) > 1$. If this is the case you can use the basic definition of a logarithm: $\displaystyle x = 10^{\log_{10}(x)}$
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  5. #5
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    Hmm I am still kind of confused on how to solve it?
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  6. #6
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    $\displaystyle 10^{\log(x^6)} > 10^{1} \rightarrow x^6 > 10$
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  7. #7
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    Hello, homeylova223!

    $\displaystyle \text{1. Use natural logarithms to solve the inequality: }\:x^{\frac{2}{3}} \:> \:27.6$

    Take logs: .$\displaystyle \ln\left(x^{\frac{2}{3}}\right) \:>\:\ln(27.6) $

    . . . . . . . . . .$\displaystyle \tfrac{2}{3}\ln(x) \:>\:\ln(27.6) $

    . . . . . . . . . . . $\displaystyle \ln(x) \:>\:\tfrac{3}{2}\ln(27.6) \:=\:4.976723659$

    . . . . . . . . . . . . . .$\displaystyle x \:>\:e^{4.976723659}$

    . . . . . . . . . . . . . .$\displaystyle x \:>\:144.9985379$




    $\displaystyle \text{2. Solve using common logarithms: }\:\log(x^6)\: > \: 1$

    We have: .$\displaystyle \log(x^6) \:>\:\log(10)$

    . . . Then: . . . . $\displaystyle x^6 \:>\:10 $

    Therefore: . . . .$\displaystyle x \:>\:\sqrt[6]{10}$

    . . . . . . . . . . . . $\displaystyle x \:>\:1.467799268$

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  8. #8
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    It is important to note that the logarithm, with any positive base, is an increasing function: if a> b then log(a)> log(b).
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