2e^7x +4=4
2e^7x=0
e^7x=0
ln1=0
so 7x=1
x=1/7
v.easy
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!!!
4 answers
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.
the e actually represect ln.
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
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
you are correct. I understand it now. You are such a fantastic help :)
1 answer