Asked by Po
Find inverse equation of
y= 3e^(x-1)^5
Answer I got is (ln(x/3))^(1/5) +5 =y
But I think it's wrong because the domain and range is really different from the original equation
y= 3e^(x-1)^5
Answer I got is (ln(x/3))^(1/5) +5 =y
But I think it's wrong because the domain and range is really different from the original equation
Answers
Answered by
Reiny
step 1: interchange the x's and y's
x = 3 e^(y-1)^5
take ln of both sides
ln x = ln3 + (y-1)^5
(y-1)^5 = lnx - ln3 = ln(x/3)
y - 1 = (ln(x/3) )^(1/5)
y = (ln(x/3) )^(1/5) + 1
check: let x = 2.2
y = 3e^2.48832 = appr 36.123
plug that into second function
y = 2.2
My answer is Not INcorrect.
x = 3 e^(y-1)^5
take ln of both sides
ln x = ln3 + (y-1)^5
(y-1)^5 = lnx - ln3 = ln(x/3)
y - 1 = (ln(x/3) )^(1/5)
y = (ln(x/3) )^(1/5) + 1
check: let x = 2.2
y = 3e^2.48832 = appr 36.123
plug that into second function
y = 2.2
My answer is Not INcorrect.
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