Question

Solve for x if e^2x+1=5

Answer 1

Get e by itself first, so move that 1

Answer 2

Bro, are you dead or something?

Answer 3

Sorry, I trying to figure this out.

Answer 4

lol and I am not a guy
ln(e^2x+1)=ln(5)

Answer 5

do what luigi suggested first.

Answer 6

cross out the ln and e

Answer 7

is it e^(2x+1) or e^(2x) +1 ??

Answer 8

is the 1 included in the exponent ?

Answer 9

Yes it is the 2x+1 is the exponent together

Answer 10

Oh x_x My apologies then.

Answer 11

cool, so take ln on both sides,

Answer 12

\(e^a=b \implies a = \ln b\)

Answer 13

I am not sure what b and a is?

Answer 14

nevermind,
take ln on both sides of e^(2x+1)=5
what do u get ?

Answer 15

2x +1= ln5

Answer 16

just subtract 1 from both sides

Answer 17

then divide both sides by 2, and you would get 'x'

Answer 18

Not to steal the answer but sometimes seeing it like this helps, good job explaining @hartnn

e^(2x+1)=5 Original Question

ln(e^(2x+1)=ln(5) Take the natural log of both sides (ln)
2x+1=ln(5) The ln cancels the e and the right side doesn't change
2x=ln(5)-1 subtract the one over and then divide by two to get:\[x=\frac{ \ln (5)-1 }{ 2 }\]

Answer 19

Thanks @hartnn and @lasttccasey very clear.
But I have one question about the part where ln and e getting canceled? How does that get canceled.

Answer 20

ln is the opposite of e functions.

Answer 21

\(\Large \ln e^y = y \ln e = y\times 1\)
as ln e =1

Answer 22

by the property that \(\log_aa=1\)

Answer 23

oh so it = 1 that is why it is canceled.

Answer 24

ln e does = 1 because its actually ln(e^1) so the power on the e becomes the remainder. log and natural logs are slightly different.

Answer 25

I see, wow thanks, the rest is clear.

Answer 26

You're welcome.