Posted by **Lee** on Sunday, November 14, 2010 at 2:33am.

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 -
**john**, Sunday, November 14, 2010 at 3:28am
2e^7x +4=4

2e^7x=0

e^7x=0

ln1=0

so 7x=1

x=1/7

v.easy

- College Algebra -
**Lee**, Sunday, November 14, 2010 at 3:45am
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 -
**Reiny**, Sunday, November 14, 2010 at 10:41am
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 -
**Lee**, Tuesday, November 16, 2010 at 10:46pm
you are correct. I understand it now. You are such a fantastic help :)

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