How do you solve this: e^7x+1 = 115, solve for x?

Answer 1

take the natural log of both sides to cancel out the e so
ln (e^7x+1) = ln (115)

7x+1= ln (115)

subtract 1 from both sides. then divide both sides by 7.

7x = ln(115)-1

x = ( ln (115) - 1) / 7

x is roughly equal to 0.535

Answer 2

Assuming the 1 is in the exponent:

e^(7x + 1) = 115

7x + 1 = ln(115)

x = (ln(115) - 1)/7 ~ 0.53499

Answer 3

Take natural log (ln) of both sides:

ln(e^7x) + ln 1 = ln 115 ---> because ln e^anything = anything
7x + ln 1 = ln 115 (because ln 1 = 0)

Now you can solve for X which I believe will be equal to
x= ln 115/7 = 0.67785

Answer 4

put natural log (Ln) in front of both sides of equation: Ln e^7x+1 = Ln 115

move exponent in front of Ln: 7x+1 Ln e = Ln 115

Ln e is equal to 1, so that just cancels out: 7x+1 = Ln 115

subtract 1 on both sides: 7x = Ln 115 - 1
*use graphing calculator for Ln 115 -1
*in graphing calculator, make sure to type this : Ln (115) - 1

after figuring out what Ln 115 - 1 equals, divide that answer by 7 and you'll get the answer for x :)

Answer 5

1. Subtract 1 from both sides.
e^7x = 114

2. Take the natural log of both sides to get rid of "e."
ln (e^7x) = ln 114

3. By log properties,
7x = ln 114

4. Divide both sides by 7 to solve for x.
x = ln 114 / 7
x is approximately .677

Answer 6

e^7x + 1 = 115

e^7x = 115-1

x = 114/(e^7)

x = 0.10395

Keep in mind that e is just another number. It's not a variable.