Results 1 to 3 of 3

Thread: Exponents as logs?

  1. #1
    Banned
    Joined
    Oct 2009
    Posts
    56

    Exponents as logs?

    x^log x = 100x

    How do I solve this? My book says the answers are 100 and .1 but how do I prove this. There is NOTHING in the book that answers this. Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,854
    Thanks
    138
    Quote Originally Posted by satx View Post
    x^log x = 100x

    How do I solve this? My book says the answers are 100 and .1 but how do I prove this. There is NOTHING in the book that answers this. Thanks
    1. $\displaystyle x = 10^{\log(x)}$

    2. Your equation becomes:

    $\displaystyle x^{\log( x)} = 100 \cdot x$

    $\displaystyle \left(10^{\log(x)}\right)^{\log(x)} = 10^2 \cdot 10^{\log(x)}$

    $\displaystyle \left(\log(x)\right)^2-\log(x)-2=0$

    3. This is a quadratic equation in $\displaystyle \log(x)$ . Solve for $\displaystyle \log(x)$. You should come out with $\displaystyle \log(x) = 2 \vee \log(x)=-1$.

    4. Calculate x.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, satx!

    Solve for $\displaystyle x\!:\;\;x^{\log x} \:=\: 100x$
    Note that $\displaystyle x \neq 0.$

    Divide by $\displaystyle x\!:\;\;x^{\log x - 1} \:=\:100$

    $\displaystyle \text{Take logs, base }x\!:\;\;\log_x\left(x^{\log x - 1}\right) \:=\:\log_x(100) \quad\Rightarrow\quad (\log x - 1)\cdot\underbrace{\log_x(x)}_{\text{This is 1}} \:=\:\log_x(100)$

    We have: . $\displaystyle \log x - 1 \:=\:\log_x100$

    Change-of-base Formula: .$\displaystyle \log x - 1 \:=\:\frac{\log 100}{\log x} \quad\Rightarrow\quad \log x - 1\:=\:\frac{2}{\log x}$

    Multiply by $\displaystyle \log x\!:\;\;(\log x)^2 - \log x \:=\:2 \quad\Rightarrow\quad (\log x)^2 - \log x - 2 \:=\:0$


    Factor: .$\displaystyle (\log x + 1)(\log x - 2) \:=\:0$

    And we have: .$\displaystyle \begin{array}{cccccccccccc}
    \log x + 1 \;=\:0 & \Rightarrow & \log x \:=\:\text{-}1 & \Rightarrow & x \:=\:10^{-1} \:=\:\boxed{\frac{1}{10}}\\ \\[-3mm]
    \log x - 2 \:=\:0 & \Rightarrow & \log x \:=\:2 & \Rightarrow & x \:=\:10^2 \:=\:\boxed{100}\end{array}$

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Exponents and Logs
    Posted in the Algebra Forum
    Replies: 2
    Last Post: Jun 2nd 2010, 06:36 PM
  2. Exponents and Logs
    Posted in the Algebra Forum
    Replies: 5
    Last Post: May 30th 2010, 05:17 PM
  3. PLS HELP...Exponents and Logs
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: Feb 26th 2010, 03:40 AM
  4. Logs and Exponents
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Jan 18th 2009, 03:35 PM
  5. exponents and logs
    Posted in the Algebra Forum
    Replies: 7
    Last Post: May 15th 2007, 01:52 PM

Search Tags


/mathhelpforum @mathhelpforum