Been trying to figure this one out for some time

[imperiale]

LogeX=2loge(100-y^2)+C where X(0)=100 find C.
i keep getting different ans, don't know where i am going wrong

 

[HallsofIvy]

We can't tell you what you are doing wrong if you don't tell us what you are doing!

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?

 

[zgozvrm]

LogeX=2loge(100-y^2)+C where X(0)=100 find C.
i keep getting different ans, don't know 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

 

[HallsofIvy]

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

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

 

[zgozvrm]

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)