How Do You Solve a Natural Logarithm Equation for Variable y?

[alextsipkis]

hello,

I was trying to figure out what will be the y value for this equations:

1 - e^-0.15*10^-5*y = 0.1

Could somebody help me in this?? The answer is supposed to be 70,240.

thanks.

 

[gel]

1 - e^-0.15*10^-5*y = 0.1

What does that mean?
1 - (e^-0.15)*(10^-5)*y = 0.1
1 - (e^-0.15)*(10^(-5*y)) = 0.1
1 - (((e^-0.15)*10)^-5)*y = 0.1
1 - e^(-0.15*(10^(-5*y))) = 0.1
1 - e^(-0.15*(10^-5)*y) = 0.1
1 - e^(-0.15*(10^(-5*y))) = 0.1

 

[alextsipkis]

Sorry , i forgot to put the parenthesis. Here , it goes.. Hope it will be clear this time.

Note that the parameter lambda has this value:

lambda = (0.15*10^-5)

And i was trying to calculate,

(1 - e^(-lambda*y)) = 0.1 for y, and the answer is assumed to be 70,240.

Similarly,

1 - e^(-lambda*y) = 0.632, and the answer for this is supposed to be 666,667.


I am really confused how we get to this answer. I am not sure wether we use log or ln to get this answer.

thanks.

 

[gel]

well, ln(x) is the inverse of the exponential function exp(x) == e^x, so you must use that. Rearrange it so that you just have e^(something) on one side of the equation, and apply ln to both sides (actually mathematicians often use log to mean ln, but calculators use ln to mean natural log, log to mean base 10 log).

 

[alextsipkis]

I tried, but not getting the right answer.

Could somebody clear this confusion...

thanks.

 

[gel]

http://tinyurl.com/r4nrzd"

 

[alextsipkis]

Thanks a lottttttttt gel...i got it right now... :-)