i know the answer is -9e^2+1/ 5e^2
my teacher didn't teach me about negative logs so i'm trying to figure it out
my teacher didn't teach me about negative logs so i'm trying to figure it out
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OK. You've been given a bunch of answers...but let's work the process.
You know the e^ln(x) = x right. and e^-2 = 1/e^2 right.
So, 5p+9 = 1/e^2 and this leads to 5p = (1/e^2)-9 or (1-9e^2)/e^2 and if you divide both sides by 5 you get the answer you provided
p = (1-9e^2)/(5e^2)
You know the e^ln(x) = x right. and e^-2 = 1/e^2 right.
So, 5p+9 = 1/e^2 and this leads to 5p = (1/e^2)-9 or (1-9e^2)/e^2 and if you divide both sides by 5 you get the answer you provided
p = (1-9e^2)/(5e^2)
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5p + 9 = e^-2 This happens when you take the antilog of a number.
5p = - 9 + e^-2 Now divide by 5
p = -9/5 + 1/5e^2 This should be your answer.
or
(-45 * e^2 + 1) / 5 e^2
JC
5p = - 9 + e^-2 Now divide by 5
p = -9/5 + 1/5e^2 This should be your answer.
or
(-45 * e^2 + 1) / 5 e^2
JC
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I think you have a typo and the negative sign is a mistake.
thus e^2=5p+9
p=(e^2-9)/5=e^2(1-9/e^2)/5 or "-9e^2+1/ 5e^2"
thus e^2=5p+9
p=(e^2-9)/5=e^2(1-9/e^2)/5 or "-9e^2+1/ 5e^2"
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You can't take the log of a negative number. Ever. So this problem is undefined.
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ln(5p+9)= -2 ----->5p+9 = e^(-2) ---> p = ...