- #1

- 2

- 0

i keep getting different ans, dunno where i am going wrong

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter imperiale
- Start date

- #1

- 2

- 0

i keep getting different ans, dunno where i am going wrong

- #2

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 965

Also, what do you mean by X(0)? Are you thinking of X as a function of y? That is, X(0)= 100 means that if y= 0, X= 100?

- #3

- 754

- 1

i keep getting different ans, dunno where i am going wrong

Assuming this equation is more correctly written as:

Log(e^X) = 2 * Log[e^(100-Y^2)] + C

and, assuming

... what do you mean by X(0)? Are you thinking of X as a function of y? That is, X(0)= 100 means that if y= 0, X= 100?

then we have:

Log(e^100) = 2 * Log(e^100) + C

and C = -Log(e^100)

But, if you really mean to use natural log (Ln), rather than the base 10 logarithm (Log), the equation becomes:

Ln(e^100) = 2 * Ln(e^100) + C

and, since Ln(e^x) = x, we have:

100 = 2 * 100 + C

100 = 200 + C

C = -100

- #4

HallsofIvy

Science Advisor

Homework Helper

- 41,847

- 965

Hopefully, imperiale will come back and explain in more detail exactly what he is trying to do.

- #5

- 754

- 1

I assumed that he does not mean log(e^{x}) but that "Loge" meant "log base e" or natural logarithm.

If this is the case, then using Log base e (better known as natural logarithm, or "Ln"), the equation becomes:

Ln(X) = 2 * Ln(100 - Y^2) + C

and, still assuming

... what do you mean by X(0)? Are you thinking of X as a function of y? That is, X(0)= 100 means that if y= 0, X= 100?

we would have:

Ln(100) = 2 * Ln(100) + C

and C = -Ln(100)

Share: