- #1

- 18

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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.

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- Thread starter alextsipkis
- Start date

- #1

- 18

- 0

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.

- #2

mathman

Science Advisor

- 8,022

- 526

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

- 18

- 0

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

- 6,531

- 1,366

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".

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]

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)

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

[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

The decimal approximation of this is about 21,072.

edit: it is actually 70,240 (I typed it wrong on my calculator)

Last edited:

- #6

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- 0

- #7

- 179

- 0

y = 70240.3 but yes, the equation is correct.

Thanks, my fault. I typed 0.5 rather than 0.15 in my calculator.

- #8

- 18

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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

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

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

[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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