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Math Help - 8 problems that I can't understand after trying for days and looking examples online

  1. #16
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    Re: 8 problems that I can't understand after trying for days and looking examples onl

    Quote Originally Posted by abender View Post
    f(x)=(x-2)^3


    y=(x-2)^3


    Switching x and y:
    x = (y-2)^3


    Now, to solve for y, raise both sides to the \tfrac{1}{3} power:
    x^{\frac{1}{3}} = ((y-2)^3)^{\frac{1}{3}}


    x^{\frac{1}{3}} = y-2 \implies y = x^{\frac{1}{3}}+2


    REMEMBER, the cube root is the same as the one-third power, just like the square root of a number is the same as the number raised to the one-half.

    Oh, I get it. The "cube root is the same as one-third power" messed me up. Way clearer now. I was trying to do it with: (y-2)(y-2)(y-2).


    Quote Originally Posted by abender View Post
    Do you mean y^{2}x?

    Yes.
    Last edited by Heisenberg; December 17th 2012 at 07:47 PM.
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  2. #17
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    Re: 8 problems that I can't understand after trying for days and looking examples onl

    Question 8:

    A simple Google search for "log properties" yields Logarithmic Properties -- LEARN ALL OF THESE!

    These properties must be understood to solve this problem:
    1) \log_b(xy) = \log_b(x) + \log_b(y)
    2) \log_b(\tfrac{x}{y}) = \log_b(x) - \log_b(y)
    3) \log_b(x^n) = n\log_b(x)

    First, change \displaystyle^8\sqrt{7} to 7^{\tfrac{1}{8}}, which I explained earlier.

    \small{\log_{15}\frac{7^{\frac{1}{8}}}{y^2x} = \color{red}\log_{15}(7^{\frac{1}{8}}) \color{black} - \color{blue}\log_{15}(y^2x)\color{black} = \color{red}\frac{1}{8}\log_{15}(7)\color{black} - \color{blue} \left(\underbrace{\log_{15}(y^2)}_{\text{Same as }2\log_{15}(y)} + \log_{15}(x) \right) \color{black}= \tfrac{1}{8}\log_{15}(7) - 2\log_{15}(y) - \log_{15}(x)}
    Thanks from Heisenberg
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  3. #18
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    Re: 8 problems that I can't understand after trying for days and looking examples onl

    abender, I am just letting you know that I took the test and I nailed it! Never did I do a test better than this one. And you helped me a lot, those problems that you helped me solve and understand, were in the test and you are the reason that I knew how to do them and do them correctly. Thank you again! Happy holidays!
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