SOLUTION: Hi, I hope you please can help me with this one;
I have found this equation from a graph I created: y = 50,88e^-0,01x
I already know that y = 50, so I have to find the x. I kn
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-> SOLUTION: Hi, I hope you please can help me with this one;
I have found this equation from a graph I created: y = 50,88e^-0,01x
I already know that y = 50, so I have to find the x. I kn
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Question 119762: Hi, I hope you please can help me with this one;
I have found this equation from a graph I created: y = 50,88e^-0,01x
I already know that y = 50, so I have to find the x. I know I have to divide/multiply with ln or something like that, but I can't come any further on my own.
I would really appriciate your help! Answer by ankor@dixie-net.com(22740) (Show Source):
You can put this solution on YOUR website! I have found this equation from a graph I created:
y = 50.88e^-0,01x
I already know that y = 50, so I have to find the x.
:
50.88e^-0.01x = 50
:
Divide both sides by 50.88
e^-0.01x =
:
Using nat logs;
ln(e^-0.01x) = ln()
:
Write the log equiv of exponents
-.01x*n(e^) = ln()
:
Find the ln of both sides (we know ln of e = 1), so we have:
-.01x = -.0174469
:
x =
:
x = +1.74469
:
:
Check solution on a good calc: enter 50.88(e^(-.01*1.74469)) = 50
:
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