SOLUTION: Hi. This is for an AP Calculus Summer Assignment
Please help
{{{ 50e^(-x)= 30 }}}
I don't even know where to start because I don't know how to handle the negative exponen
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Question 766211: Hi. This is for an AP Calculus Summer Assignment
Please help
I don't even know where to start because I don't know how to handle the negative exponent
Thank you
Found 3 solutions by solver91311, kmadison, rothauserc:
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
But that doesn't really come into play in this problem.
Multiply both sides by 
Take the natural log of both sides
\ =\ \ln\left(\frac{3}{5}\right))
Then use
\ =\ n\log_b(x))
and
\ =\ 1)
To write
)
Then use
 - \log_b(y) = \log_b\left(\frac{x}{y}\right))
to write
\ -\ \ln(5))
Then multiply by -1
\ -\ \ln(3))
Your instructor might also be looking for
)
or
)
Any of the three forms is correct.
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
Answer by kmadison(20) (Show Source): You can put this solution on YOUR website!
First divide both sides by 50:
Now take the natural log of both sides:
Finally, divide by -1 to get your final answer:
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
50e^-x= 30
50 / e^x = 30 here apply exponent rule 4^-3 = 1/4^3
multiply both sides of = by e^x
30*e^x = 50
e^x = 50/30 = 5/3
log(e)5/3 = x this is log base e of 5/3 = x
ln 5/3 = x ln is the natural log
x = .51
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