Question

Does anyone know what to do?
1/ex=2

Answer 1

\[\frac{ 1 }{ ex }=2 \]

Answer 2

\(\Large \frac{1}{e^x} = 2\) ?

Answer 3

yes

Answer 4

\(\Large \frac{1}{e^x} = 2\)
\(\Large e^{-x} = 2\)
Take natural logarithm on both sides:
\(\large \ln(e^{-x}) = \ln(2)\)
\(\large -x * \ln(e) = \ln(2)\)
\(\large -x * 1 = \ln(2)\)
\(\large x = -\ln(2)\)

Answer 5

OR

\(\large x = -\ln(2) = \ln(2)^{-1} = \ln(1/2) = \ln(0.5)\)

Answer 6

ok ao since it's a fraction we need to get rid of the fraction by converting it to e(exponent -x)

Answer 7

Thank you very much for your help! :)

Answer 8

It becomes easier to solve. You can also take natural logarithm of the first line:
ln(1/e^x) = ln(2)
ln(A/B) = ln(A) - ln(B)
ln(1/e^x) = ln(1) - ln(e^x) = ln(2)
ln(1) = 0
ln(e^x) = x.
So 0 - x = ln(2)
x = -ln(2)

Answer 9

the answer i got is x=-0.6931471806

Answer 10

correct. Round it to whatever decimal places they tell you to.
But if none specified, I will leave the answer as x = ln(0.5) or x = -ln(2).

Answer 11

oh ok the first one was easier to understand than the second one. Thank you very much!

Answer 12

You are welcome.

Answer 13

I have another question, just so I can see if I understood the problem. I have another problem \[\ln (2x)=7\]\[x \times \ln (2)=\ln (7)\]\[x \times0.69=1.94\]\[\frac{ 0.69 }{ 0.69 }x=\frac{ 1.94 }{ 0.69 }\]\[x=2.811\]
am i right or wrong?

Answer 14

\(\ln(AB) = \ln(A) + \ln(B)\)

Answer 15

\(\ln(2x) = 7\) implies \(2x = e^7; ~~x = \frac 12 e^7\)

Answer 16

\(\ln(2x) \ne x \times \ln(2)\)

\(\ln(2x) = \ln(2) + \ln(x)\)

Answer 17

Or is it \(2^x\) and not \(2x\) ?

Answer 18

not exponent, it's 2x

Answer 19

Ok, then what I said above is correct.

\(\ln(2x) = \ln(2) + \ln(x)\)

\(\ln(2^x) = x \times \ln(2) \)

Answer 20

thank you !

Answer 21

you are welcome.