exponential

Hey

e^{2-2x} = 2e^{-x}

is this correct:

lne^{2-2x} = ln2e^{-x}[br](2-2x)(lne) = -x(ln2e)[br](2-2x) = -x(lne)+(ln2)[br]2-2x = -x + ln2[br]-2x = -x + ln2 -2[br]2x = x - ln2 + 2[br]x = 2 - ln2

(edited 13 years ago)

Nope. Your error is going from

\ln (2e^{-x})

-x \ln (2e)

. Notice that

-x\ln (2e) = \ln (2e)^{-x} = \ln (2^{-x}e^{-x})

, and not

\ln (2e^{-x})

Nope. Your error is going from

\ln (2e^{-x})

-x \ln (2e)

. Notice that

-x\ln (2e) = \ln (2e)^{-x} = \ln (2^{-x}e^{-x})

, and not

\ln (2e^{-x})

Thank! how about now

lne^{2-2x} = ln2e^{-x}[br](2-2x)(lne) = (ln2) + (lne^-x)[br](2-2x) = (ln2) + (-x)(lne)[br]2-2x = ln2 - x[br]-2x = ln2 - x - 2[br]2x = -ln2 + x +2[br]x = -ln2 + 2

(edited 13 years ago)

Original post by Limoncello

Thank! how about now

lne^{2-2x} = ln2e^{-x}[br](2-2x)(lne) = (ln2) + (lne^-x)[br](2-2x) = (ln2) + (-x)(lne)[br]2-2x = ln2 - x[br]-2x = ln2 - x - 2[br]2x = -ln2 + x +2[br]x = -ln2 + 2

That's right.

Original post by Limoncello

Thank! how about now

lne^{2-2x} = ln2e^{-x}[br](2-2x)(lne) = (ln2) + (lne^-x)[br](2-2x) = (ln2) + (-x)(lne)[br]2-2x = ln2 - x[br]-2x = ln2 - x - 2[br]2x = -ln2 + x +2[br]x = -ln2 + 2

That's right :smile:

In questions like this you can check your answer by substituting it into both sides of the given equation to see if they're equal.

Original post by Farhan.Hanif93

That's right.

Thanks

LaTex

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