Results 1 to 3 of 3

Math Help - Exponential functions help

  1. #1
    Newbie
    Joined
    Nov 2011
    Posts
    1

    Exponential functions help

    Basically, I don't get exponential equations. Could someone please help me? These are the two I'm having the most trouble with:

    2^(1 - 2x) = 3x

    (7/9)^x=11^1-x
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Siron's Avatar
    Joined
    Jul 2011
    From
    Norway
    Posts
    1,250
    Thanks
    20

    Re: Exponential functions help

    To solve the second equation you can write:
    \left(\frac{7}{9}\right)^x=11\cdot 11^{-x}
    \Leftrightarrow \left(\frac{7}{9}\right)^x=\frac{11}{11^{x}}
    \Leftrightarrow \left(\frac{7}{9}\right)^x \cdot 11^x=11

    (You can also take the logarithm of both side in the beginning and continue)

    Now use (the general rule) a^{x}\cdot b^{x}=(a\cdot b)^x

    About the first equation, are you sure it isn't 2^{1-2x}=3^x in stead of 2^{1-2x}=3x?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,373
    Thanks
    1314

    Re: Exponential functions help

    Quote Originally Posted by hotpotato1092 View Post
    Basically, I don't get exponential equations. Could someone please help me? These are the two I'm having the most trouble with:

    2^(1 - 2x) = 3x

    (7/9)^x=11^1-x
    Do you understand that the logarithm is the inverse function to the exponential? And, it turns out you can use logarithm to any base!

    You also need to learn the "rules of expentials":
    a^x a^y= a^(x+ y) and (a^x)^y= a^(xy)
    and the corresponding "rules of exponentials:
    log(ab)= log(a)+ log(b) and log(a^b)= b log(a).
    It is that last one that is of the most use.

    Taking the logarithm of both sides of that second equation, as Siron suggests, gives
    log((7/9)^x)= x log(7/9)= (1-x)log(11). You now have a linear equation to solve for x.

    I agree with Siron also that you should look again at the first problem. Although that can be solve, I doubt it is what is intended.

    2^(1- 2x)= 2(2^(-2x))= 2(1/4)^x= 3x. dividing both sides by 3 and Multiplying both sides by (1/4)^x you get x4^x= 2/3. Of course, 4= e^ln(4) so 4^x= e^(xln(4)) and the equation is xe^(xln(4))= 2/3. Now multiply both sides by ln(4) so the equation is (xln(4))e^(xln(4))= 2ln(4)/3. Almost there! Let y= x ln(4). The equation ye^y= 2ln(4)/3 has solution y= W(2ln(4)/3) where "W(x)" is the "Lambert W function"- which is defined as the inverse function to f(x)= xe^x. Since y= W(2ln(4)/3) and y= x ln(4), x= W(2ln4)/3)/ln(4).

    But that is NOT the sort of problem anyone who is not in a "special functions" course where they are introduced to the W function would be expected to do!

    If the problem is in fact 2^(2x-1)= 3^x, take the logarithm of both sides:
    (1- 2x)log(2)= x log(3). Again, you have a linear equation to solve for x.
    Last edited by HallsofIvy; November 21st 2011 at 08:32 AM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: June 9th 2011, 09:54 AM
  2. Exponential functions
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: December 7th 2008, 04:26 PM
  3. Exponential Functions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: May 12th 2008, 08:50 PM
  4. Exponential Functions
    Posted in the Algebra Forum
    Replies: 5
    Last Post: April 23rd 2008, 01:20 PM
  5. exponential functions..
    Posted in the Algebra Forum
    Replies: 4
    Last Post: June 3rd 2007, 10:50 AM

Search Tags


/mathhelpforum @mathhelpforum