Can somebody solve this equation for y using ln or any other technique

  • #1
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.
 

Answers and Replies

  • #2
mathman
Science Advisor
7,890
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Without paraentheses there is some ambiguity. I'll assume you mean:
1 - (e^-0.15)*(10^-5*y) = 0.1
Then (e^-0.15)*(10^-5*y) = 0.9
log(e^-0.15) + log*(10^-5*y) = log(.9)
I'll use base 10 logs
-.15log(e) - 5y = log(.9)
You can finish.
However I don't think you will get the answer you gave. Check where the parentheses belong.
 
  • #3
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. Plz helpppppppppp!
 
  • #4
symbolipoint
Homework Helper
Education Advisor
Gold Member
6,104
1,152
Let me give an uncalculated but numeric answer without the steps to the first question:

y = (-ln(0.90))/(0.15*10^(-5))

Check your steps to see if you find that same or equivalent expression.

My attempt to use the formatting tool -

[tex]\frac{-ln(0.90)}{0.15\times10^{-5}}[/tex]
Excuse me if that does not work, and refer to the first expression at "y = ..."
Yes, seems what I wanted to show (the TEX version)
EDIT: the forum needs tags for "strike-through".
 
Last edited:
  • #5
179
0
[tex]1-e^{-0.15*10^{-5}*y}=0.1[/tex]

What you need to do is make y the subject. The 'y' is currently inside the exponential (e^), and the way you can get rid of the e^ is to ln() it. Because ln(e^x) = x.

First you take 1 from each side, and then make both sides positive (multiply by -1).

[tex]e^{-0.15*10^{-5}*y}=0.9[/tex]

Then you can log both sides with the natural logarithm:

[tex]\ln{(e^{-0.15*10^{-5}*y})}=\ln{0.9}[/tex]

And because ln(e^x) = x (this is just the definition of the natural log), you can get rid of the ln() and the e^ on the left, because they cancel each other out.

[tex]-0.15*10^{-5}*y=\ln{0.9}[/tex]

You then divide both sides by (-0.15*10^(-5)):

[tex]\frac{(-0.15*10^{-5})*y}{(-0.15*10^{-5})}=\frac{\ln{0.9}}{-0.15*10^{-5}}[/tex]

[tex]y=\frac{\ln{0.9}}{-0.15*10^{-5}}[/tex]

This is the same result as symbolipoint got.

The decimal approximation of this is about 21,072.

edit: it is actually 70,240 (I typed it wrong on my calculator) :redface:
 
Last edited:
  • #8
Thanks for all your replies...especially Georgepowell, whose step by step solving really helped me understand it well.

Could somebody also help me in taking the partial derivate of the following expression with respect to lambda.

(xln(lambda))

thanks a lot.
 
  • #9
179
0
Could somebody also help me in taking the partial derivate of the following expression with respect to lambda.

(xln(lambda))
Because it is the partial derivative with respect to lambda, all variables other than lambda are presumed to be constant.

The derivative of ln(lambda) is 1/lambda. And because x is a constant, the derivative of x*ln(lambda) is just x*1/lambda = x/lambda.

Here is an explanation of why the derivative of ln(lambda) is 1/lambda:

[tex]y = \ln(x)[/tex]

[tex]x = e^{y}[/tex] ...Because e^() is the inverse of ln()

[tex]\frac{dx}{dy} = e^{y}[/tex] ...This is just a property of e^(), its derivitive is also e^()

[tex]\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{e^{y}} = \frac{1}{x}[/tex] ...because x = e^y, as stated in the second line
 
Last edited:

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