Asked by Jon
I need to find the future population using the model P=42+20ln(11t+1)
t= time in years
I need to find when it will be 180. I have my work below, but I am stuck. The answer I get is not right, it will not add up to 180. Maybe I am wrong some where.
P=42+20ln(11t+1)
180=42+20ln(11t+1)
-42 -42
(138/20)=(201n(11t+1)/20)
6.9=1n(11t+1)
e(6.9)=e^ln(11t+1)
e(6.9)=11t+1
-1 -1
[e(6.9)-1/11]=(11t/11)
t=[e(6.9)-1/11]
my wrong answer: 1.614193818
t= time in years
I need to find when it will be 180. I have my work below, but I am stuck. The answer I get is not right, it will not add up to 180. Maybe I am wrong some where.
P=42+20ln(11t+1)
180=42+20ln(11t+1)
-42 -42
(138/20)=(201n(11t+1)/20)
6.9=1n(11t+1)
e(6.9)=e^ln(11t+1)
e(6.9)=11t+1
-1 -1
[e(6.9)-1/11]=(11t/11)
t=[e(6.9)-1/11]
my wrong answer: 1.614193818
Answers
Answered by
bobpursley
When you divide by 11, don't you need to divide e^6.9 by 11 also?
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