Question 236978: 2= 5e^-0.35h
Your are trying to find H
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! 2 = 5e^-0.35h
divide both sides of the equation by 5 to get:
2/5 = e^(-.35h)
take natural log of both sides to get:
ln(2/5) = ln(e^(-.35h)
since log(b^x) = x*log(b), your equation becomes:
ln(2/5) = -.35h * (ln(e))
since ln(e) = 1, your equation becomes:
ln(2/5) = -.35h
divide both sides by -.35 to get:
ln(2/5)/(-.35) = h
since ln(2/5) = (-.916290732), your equation becomes:
(-.916290732)/(-.35) = h
simplify to get:
h = 2.61797352
to prove this is correct, replace h in the original equation to get:
2/5 = e^(-.35h) becomes:
2/5 = e^(-.35*2.61797352)
this becomes:
2/5 = e^(-.916290732)
this becomes:
.4 = .4 confirming the answer is correct.
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