Question

2e^x-1=9

Answer 1

\[\large 2e^{x-1}=9\]\[\large e^{x-1} =\frac{9}{2}\]\[\large \ln e^{x-1} =\ln \frac{9}{2}\]\[\large x-1= \ln (9) -\ln (2)\]...

Answer 2

@bahrom7893

Answer 3

uhm was -1 in the power?

Answer 4

no it's minus one

Answer 5

So add 1 to both sides first.

Answer 6

If not.... \[\large 2e^x -1 = 9\]\[\large 2e^x =10\]\[\large e^x = 5\]\[\ln e^x =\ln 5\]\[x= ?\]

Answer 7

thanksss u guys=]] u helped alot

Answer 8

I didn't know, you didnt indicate in your problem whether -1 was or was not included in the power, so it became a little confusing.

Answer 9

would the answer come out as a decimal

Answer 10

Nope.

Answer 11

would it be ln 5

Answer 12

you should know that ln e^{x} = x. natural log has a base of e, so it's technically saying \(\large \ln_e e^{x}\) which equals \(x\)

Answer 13

Yes, it is \(\ln 5\) good job.

Answer 14

this is really hard for me, but thank u I understand now

Answer 15

If you have e^(anything) the best way to get rid of it is by taking the natural log of both sides. Just remember that. \(\large \ln e^x = x\) and \(\large e^{\ln x} = x \)

Answer 16

ooh okkkk now i got it. Thanks

Answer 17

No problem, good luck