[MOVED] 100=39.468-(0.41352(Ln(T(100)))+(190.6/Ln(T(100)))

[bendelp]

I am trying to solve this problem for T, but can't get T by itself. I've been out of school for a while, and it's showing!

100=39.468-(0.41352(Ln(T(100)))+(190.6/Ln(T(100)))

How would this look with T by itself?

 

[stapel]

You have an unpaired parentheses somewhere; there are seven open-paren characters, but only six close-parens. Also, is there some reason you formatted as "ln(T(100))" instead of just "ln(100T)"? And is that second natural log supposed to be in the denominator?

Thank you!

Eliz.

 

[bendelp]

The original equation I am working with will find temperature (TB) when I know Time (T). I now know Temperature, and want to find Time (T).
Here is the origional equation: (x=multiply)

TB = 39.468 - 0.41352 x Ln(T x 1000) + 190.6 / Ln(T x 1000)

I am using 100 as Temperature (TB).

 

[Denis]

TB = 39.468 - 0.41352 x Ln(T x 1000) + 190.6 / Ln(T x 1000)
I am using 100 as Temperature (TB).

Next time, use * as multiplication sign.

let K = Ln(1000T)

100 = 39.468 - 0.41352K + 190.6 / K
Multiply through by K
100K = 39.468K - .41352K^2 + 190.6
.41352K^2 + 60.532K - 190.6 = 0

Solve that quadratic, then substitute back Ln(1000T)