# 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

Algebra ->  Algebra  -> Logarithm Solvers, Trainers and Word Problems -> 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      Log On

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 Click here to see ALL problems on logarithm 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(15746)   (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 : Did this help?