- #1

- 2

- 0

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

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- Thread starter imperiale
- Start date

- #1

- 2

- 0

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

- #2

HallsofIvy

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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

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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)

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