Results 1 to 3 of 3

Math Help - Exponential equations

  1. #1
    Junior Member
    Joined
    Oct 2007
    Posts
    44

    Exponential equations

    Sorry for the repost, but I missed up when I typed the problem in.

    Solve the exponential equation. Express the solution set in terms of natural logarithms.

    4^(x + 4) = 52^(x + 5 )

    Solve the exponential equation. Use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution.

    7^x = 6^(x + 7 )
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2007
    Posts
    178
    Use the power (power is in exponent, not as in power-up) rule.
    I believe that is the proper name.

    (x + 4)^4 = (x + 5 )^ 52 -huh. LaTex doesn't do exponents that big.
    (x + 4) ln 4 = (x + 5) ln 4

    \frac {(x+4) ln 4}{(x+ 5) ln 52} = 0 .... or this flipped. It's the same thing I believe.

    I believe that is what they are looking for.
    I would want a second opinion if I were you.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,715
    Thanks
    633
    Hello, soly_sol!

    Solve the exponential equation. Express the solution set in terms of natural logarithms.

    4^{x + 4} \:= \:52^{x + 5}
    Take logs: . \ln\left(4^{x+4}\right) \:=\:\ln\left(52^{x+5}\right)\quad\Rightarrow\quad  (x+4)\ln(4) \:=\:(x+5)\ln(52)

    Expand: . x\!\cdot\!\ln(4) + 4\!\cdot\!\ln(4)\:=\:x\!\cdot\!\ln(52) + 5\!\cdot\!\ln(52)

    Rearrange terms: . x!\cdot\!\ln(4) - x\!\cdot\!\ln(5) \:=\:5\!\cdot\ln(52) - 4\!\cdot\!\ln(4)

    Factor: . x\left[\ln(4)-\ln(5)\right] \:=\:5\!\cdot\ln(52) - 4\!\cdot\!\ln(4)

    Therefore: . x \;=\;\frac{5\!\cdot\!\ln(52) - 4\!\cdot\!\ln(4)}{\ln(4) - \ln(5)}




    Solve the exponential equation.
    Use a calculator to obtain a decimal approximation, to two decimal places.

    7^x \:= \:6^{x + 7}
    Take logs: . \ln\left(7^x\right)\:=\:\ln\left(6^{x+7}\right) \quad\Rightarrow\quad x\!\cdot\!\ln(7) \:=\:(x+7)\!\cdot\!\ln(6)

    Expand: . x\!\cdot\!\ln(7) \:=\:x\!\cdot\!\ln(6) + 7\!\cdot\!\ln(6)

    Rearrange: . x\!\cdot\!\ln(7) - x\!\cdot\!\ln(6)\:=\:7\!\cdot\!\ln(6)

    Factor: . x\left[\ln(7) - \ln(6)\right] \:=\:7\!\cdot\!\ln(6)

    Therefore: . x \;=\;\frac{7\!\cdot\!\ln(6)}{\ln(7)-\ln(6)} \;\approx\; 81.36

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Exponential Equations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 7th 2010, 10:40 AM
  2. Exponential equations.
    Posted in the Algebra Forum
    Replies: 2
    Last Post: September 20th 2009, 11:14 AM
  3. Exponential Equations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 15th 2009, 04:02 PM
  4. Exponential Equations
    Posted in the Calculus Forum
    Replies: 2
    Last Post: March 11th 2009, 04:19 PM
  5. exponential equations
    Posted in the Algebra Forum
    Replies: 7
    Last Post: November 11th 2007, 01:36 AM

Search Tags


/mathhelpforum @mathhelpforum