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Math Help - Problems with logaritm

  1. #1
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    Problems with logaritm

    Hello everyone! I have a few problem regarding Log.
    First of all, it is a homework about Logarithm. I provide an image of my homework here:

    (The ones squared with red are the ones I have problem with solving)
    The main problem I have with Logs are the exponents. I know how to convert from logarithm form to exponent form and vice versa. But when solving for a variable exponent, it becomes difficult.
    This is what I have done:

    In problem 57.
    I can solve a.
    But b, c, and d are different.
    In b. when solving on exponential form, how can I make a base with a variable exponent equal to a fraction (or decimal)?
    In c. when solving on exponential form, how can I make a base with a variable exponent equal to a root?
    The same goes with d.
    In problem 58.
    I could solve a. But it seems that b, c, and d, are solved differently. All of them contain negative exponents. How can I make a base with an exponent equal to another positive number if the exponent is negative?
    Thanks in advance for the help!
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  2. #2
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    Quote Originally Posted by blubla View Post
    Hello everyone! I have a few problem regarding Log.
    First of all, it is a homework about Logarithm. I provide an image of my homework here:
    ...
    (The ones squared with red are the ones I have problem with solving)
    The main problem I have with Logs are the exponents. I know how to convert from logarithm form to exponent form and vice versa. But when solving for a variable exponent, it becomes difficult.
    This is what I have done:
    ...
    to b):

    \log_2\left(\dfrac1{16}\right)=\log_2\left(\dfrac1  {2^4}\right)=\log_2\left(2^{-4}\right)=-4

    The other two problems have to be done similarly.

    to c):

    \log_2(\sqrt{8})=\log_2(\sqrt{2^3})=\log_2((2^3)^{  \frac12})=\log_2(2^{\frac32})=\dfrac32

    The other two problems have to be done similarly.

    to d):

    \log_{\frac12}(32)=\log_{\frac12}(2^5)=\log_{\frac  12}(((2^{-1})^{-1})^5)=\log_{\frac12}\left(\left(\dfrac12\right)^{-5}\right) = -5

    The other two problems have to be done similarly.
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  3. #3
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    Quote Originally Posted by blubla View Post
    Hello everyone! I have a few problem regarding Log.
    ...
    In problem 57.
    I can solve a.
    But b, c, and d are different.
    In b. when solving on exponential form, how can I make a base with a variable exponent equal to a fraction (or decimal)?
    In c. when solving on exponential form, how can I make a base with a variable exponent equal to a root?
    The same goes with d.
    In problem 58.
    I could solve a. But it seems that b, c, and d, are solved differently. All of them contain negative exponents. How can I make a base with an exponent equal to another positive number if the exponent is negative?
    Thanks in advance for the help!
    If you have the equation

    x^a = c ....... which you want to solve for x you only have to take both sides to the power \dfrac1a

    That means: x^a = c~\implies~\left(x^a\right)^{\frac1a} = c^{\frac1a}~\implies~x^1=c^{\frac1a}

    to 58 b):

    x^{-3} = 2~\implies~\left(x^{-3}\right)^{-\frac13} = 2^{-\frac13}~\implies~x=\dfrac1{\sqrt[3]{2}}

    to 58 c):

    x^{-8} = \dfrac1{16} ~\implies x^{-8} = \dfrac1{2^4} \implies x^{-8} = 2^{-4} \implies  \left(x^{-8}\right)^{-\frac18} = \left(2^{-4}\right)^{-\frac18} \implies x=2^{\frac12} \implies x=\sqrt{2}

    to 58 d):

    Keep in mind that 27 = 3. Therefore the solution is x = \dfrac13
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  4. #4
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    That helped me a lot. Thanks!
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