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Thread: Need help with Logarithm and Inverse Function

  1. #1
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    Need help with Logarithm and Inverse Function

    I am having difficulties understanding how to solve the following problems:

    1) Express as a single logarithm and, if possible, simplify:

    $\displaystyle \frac{3}{2}ln4x^6 - \frac{4}{5}ln2y^{10}$

    The book shows the answer as: $\displaystyle ln\frac{2^{11/5}x^9}{y^8}$

    I understand how the $\displaystyle x^6$ turns into $\displaystyle x^9$, but how does the $\displaystyle 4$ in $\displaystyle \frac{3}{2}ln4x^6$ turn into $\displaystyle 2^{11/5}$?

    2) Express as a single logarithm:

    $\displaystyle 7lnx + 3(lny^2 - lnz^3)$

    Since #1 is similar to #2, I am unsure on how to correctly solve this problem.

    3) Find the inverse function of:

    $\displaystyle f(x) = \frac{3x}{x-3}$

    When I do the math, it shows that the inverse function is exactly the same as the original function? How is that possible?

    I would appreciate it greatly if you could answer the questions above step-by-step.

    Thanks
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  2. #2
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    Quote Originally Posted by Ineedhelpwithlogs View Post
    I am having difficulties understanding how to solve the following problems:

    1) Express as a single logarithm and, if possible, simplify:

    $\displaystyle \frac{3}{2}ln4x^6 - \frac{4}{5}ln2y^{10}$

    The book shows the answer as: $\displaystyle ln\frac{2^{11/5}x^9}{y^8}$

    I understand how the $\displaystyle x^6$ turns into $\displaystyle x^9$, but how does the $\displaystyle 4$ in $\displaystyle \frac{3}{2}ln4x^6$ turn into $\displaystyle 2^{11/5}$?
    $\displaystyle \frac{3}{2}\ln(4x^6) - \frac{4}{5}\ln(2y^{10})$

    $\displaystyle \ln(4x^6)^{\frac{3}{2}} - \ln(2y^{10})^{\frac{4}{5}}$

    $\displaystyle \ln\left(\frac{(4x^6)^{\frac{3}{2}}}{(2y^{10})^{\f rac{4}{5}}}\right)$

    Now simplify
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  3. #3
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    Quote Originally Posted by Ineedhelpwithlogs View Post
    3) Find the inverse function of:

    $\displaystyle f(x) = \frac{3x}{x-3}$

    When I do the math, it shows that the inverse function is exactly the same as the original function? How is that possible?

    I would appreciate it greatly if you could answer the questions above step-by-step.

    Thanks
    $\displaystyle y = f(x) = \frac{3x}{x-3}$

    swap x and y and then solve for y

    $\displaystyle x = \frac{3y}{y-3}$

    $\displaystyle (y-3)x = 3y$

    $\displaystyle yx-3x = 3y$

    $\displaystyle -3x = 3y-yx$

    $\displaystyle -3x = y(3-x)$

    Can you solve for y?
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  4. #4
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    Quote Originally Posted by pickslides View Post
    $\displaystyle \frac{3}{2}\ln(4x^6) - \frac{4}{5}\ln(2y^{10})$

    $\displaystyle \ln(4x^6)^{\frac{3}{2}} - \ln(2y^{10})^{\frac{4}{5}}$

    $\displaystyle \ln\left(\frac{(4x^6)^{\frac{3}{2}}}{(2y^{10})^{\f rac{4}{5}}}\right)$

    Now simplify
    Simplifying is where I get lost - sorry, I should have mentioned that.
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  5. #5
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    Quote Originally Posted by pickslides View Post
    $\displaystyle y = f(x) = \frac{3x}{x-3}$

    swap x and y and then solve for y

    $\displaystyle x = \frac{3y}{y-3}$

    $\displaystyle (y-3)x = 3y$

    $\displaystyle yx-3x = 3y$

    $\displaystyle -3x = 3y-yx$

    $\displaystyle -3x = y(3-x)$

    Can you solve for y?
    $\displaystyle y = \frac{-3x}{3 - x}$
    $\displaystyle f^{-1}(x) = \frac{-3x}{3 - x}$

    Thank you, pickslides. I see where I made my mistake.
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  6. #6
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    $\displaystyle \frac{3}{2}\ln(4x^6) - \frac{4}{5}\ln(2y^{10})$

    $\displaystyle \ln(4x^6)^{\frac{3}{2}} - \ln(2y^{10})^{\frac{4}{5}}$

    $\displaystyle \ln\left(\frac{(4x^6)^{\frac{3}{2}}}{(2y^{10})^{\f rac{4}{5}}}\right)$

    $\displaystyle \ln\left(\frac{4^{\frac{3}{2}}x^{6\times \frac{3}{2}}}{2^{\frac{4}{5}}y^{10\times\frac{4}{5 }}}\right)$
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  7. #7
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    Quote Originally Posted by pickslides View Post
    $\displaystyle \frac{3}{2}\ln(4x^6) - \frac{4}{5}\ln(2y^{10})$

    $\displaystyle \ln(4x^6)^{\frac{3}{2}} - \ln(2y^{10})^{\frac{4}{5}}$

    $\displaystyle \ln\left(\frac{(4x^6)^{\frac{3}{2}}}{(2y^{10})^{\f rac{4}{5}}}\right)$

    $\displaystyle \ln\left(\frac{4^{\frac{3}{2}}x^{6\times \frac{3}{2}}}{2^{\frac{4}{5}}y^{10\times\frac{4}{5 }}}\right)$

    $\displaystyle \ln\left(\frac{4^{\frac{3}{2}}x^{9}}{2^{\frac{4}{5 }}y^{8}}\right)$

    then divide 4 by 2 to get:

    $\displaystyle \ln\left(\frac{2^{\frac{3}{2}}x^{9}}{{\frac{4}{5}} y^{8}}\right)$

    Now, I don't know what to do next (and I'm not sure if that last step was correct).
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