1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Point of Intersection involving logarithms

  1. Jun 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Solve for the point of intersection between ##y=\log_{2}{(2x)}## and ##y=\log_{4}{(x)}##

    2. Relevant equations
    3. The attempt at a solution

    Setting the two equations equal:
    $$\log_{2}{(2x)}=\log_{4}{(x)}\\
    2x=2^{\log_{4}{(x)}} \\
    2^{2x}=4^{\log_{4}{(x)}}\\
    2x=2\log_{4}{(x)}\\
    x=\log_{4}{(x)} \\
    4^x=x \\ $$
    I'm really confused and I'm pretty sure I've done something wrong at this point.
     
  2. jcsd
  3. Jun 11, 2015 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    You're right about that.

    If you use each side of
    ##\displaystyle \ 2x=2^{\log_{4}{(x)}}\ ##​
    as the exponent with a base of 2, you get
    ##\displaystyle \ 2^{2x}=2^{\left(\displaystyle {2^{\log_{4}{(x)}}}\right)}\ ##​

    Furthermore,
    ##\displaystyle \ \left({2^{2}}\right)^{\log_{4}{(x)}}\ne2^{\left(\displaystyle {2^{\log_{4}{(x)}}}\right)}\ ##​
     
  4. Jun 11, 2015 #3
    Managed to get the correct answer after rewriting ##2## as ##4^{1/2}## like you mentioned. $$2x=2^{\log_{4}{(x)}} \\
    2^{2x}=2^{2^{\log_{4}{(x)}}}\\
    2^{2x}=2^{4^{1/2 \cdot \log_{4}{(x)}}}\\
    2^{2x}=2^{x^{1/2}} \\
    2x-\sqrt{x}=0 \\
    \sqrt{x}(2\sqrt{x}-1)=0 $$ Giving x=0 and x=1/4, with x=0 not being a solution since it's not in the domain, thanks for the help!
     
  5. Jun 11, 2015 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    A suggestion to make things a bit easier is:

    Rather than using 2 as the base, use 4 as the base. Then change the base on the left hand side to 22 .

    ##\ \left(2^2\right)^{\log_{2}{(2x)}} =2^{\log_{2}{(4x^2)}}=\,? \ ##
     
  6. Jun 11, 2015 #5
    Okay it was definitely easier to do it this way.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Point of Intersection involving logarithms
  1. Points of intersection (Replies: 3)

Loading...