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College Algebra

posted by on .

Problem: Use Common or Natural Logarithims to solve the exponential equations symboliclly.

2e^7x+4=4
(Round to four decimal places)

I tried solving this way

2e/2^7x+4=4/2
e^7x+4=2
ln e^7x+4=ln 2
7x+4=ln 2
7x+4-4=ln 2-4
7x=ln 2-4
7x/7=ln 2-4/7
x=ln 2-4/7
then I at a lose at this point.
What step should I perform next.
I know I'm suppose to use the calculator for the estimated solution, but I'm not coming up with the correct answer. Help!!!

  • College Algebra - ,

    2e^7x +4=4
    2e^7x=0
    e^7x=0
    ln1=0
    so 7x=1
    x=1/7
    v.easy

  • College Algebra - ,

    Yes this would be very simple, except this is a logarithim equation. The answer you gave isn't correct either.

    the e actually represect ln.

  • LEE - College Algebra - ,

    Lee, the solution that John gave you is correct according to the way you typed the question.

    I have a feeling you meant to type

    2e^(7x+4) = 4

    if so, then

    e^(7x+4) = 2
    take ln of both sides
    ln(e^(7x+4)) = ln2
    (7x+4)lne = ln2 , but lne = 1
    7x+4 = ln2
    7x = ln2 - 4
    x = (ln2 - 4)/7 or appr. - .4724

  • College Algebra - ,

    you are correct. I understand it now. You are such a fantastic help :)

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