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exponential equation-Algebra

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

4^(x + 4) = 5^(2x + 5)-This question I don't understand.

  • exponential equation-Algebra -

    Alright. I think I might have gotten the answer. Write back and tell me if you think it's right. If you think I messed up, I'll keep trying.

    First use ln to rearrange the original:

    4^(x+4)=5^(2x+5)
    ln4^(x+4)=ln5^(2x+5)
    Put the exponents in front because of the properties of logs.
    (x+4)(ln4)=(2x+5)(ln5)
    Rearrange to get all the x's on one side.
    (x+4)/(2x+5)=(ln5)/(ln4)
    I'm sure you got to this point. But after here, it starts to get a little tricky.
    Technically, (ln5)/(ln4) is equal to [(ln5)/(ln4)]/1 right?
    Let's set a variable......how about "u" equal to the numerator of this fraction.
    So... u=[(ln5)/(ln4)]
    Let's plug that back in to what we had from before:
    (x+4)/(2x+5)=u/1
    Use cross multlipication to get this:
    (x+4)=(2x+5)(u)
    Distribute the u
    x+4=2xu+5u
    Get all the parts with x's on one side:
    x-2xu=5u-4
    Pull out an x:
    x(1-2u)=5u-4
    Divide the (5u-4) by (1-2u) to get x by itself:
    x=(5u-4)/(1-2u)
    Plug the u value back in and simplify
    x=(5((ln5)/(ln4))-4) / (1-2((ln5)/(ln4))
    I think that's as far as you can go.

    Good luck! :-)

  • exponential equation-Algebra -

    the closest answers that I have that are given in the multiple choice is {5 ln 5 - 4 ln 4/ln 4-2 ln 5}

  • exponential equation-Algebra -

    Take the natural logs (ln) of both sides.
    (x+4) ln 4 = (2x+5) ln 5
    x ln 4 + 4 ln 4 = 2x ln 5 + 5 ln 5
    x (2 ln 5 - ln 4) = 4 ln 4 - 5 ln 5
    x = (4ln4 - 5ln5)/(2ln5 - ln4)
    This becomes the same as your answer if you multiply both numerator and denominiator by -1

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