# How i can find X? (V*K*C)/F =(1+E)*(ln 1/1-X)-E*X?

Can anyone solve this problem?

I wrote the equation wrong, it's:

(V*K*C)/F = (1+E)*(ln(1/1-X))-E*X, and

E=constant not exponent.

### 2 Answers

- 1 decade agoFavorite Answer
First off, note that ln(1) is an actual constant and simplifies to 0. The problem is now reduced to

(VKC)/F=(1+E)*0-EX

So now, you have a variable multiplied by X. To get rid of it, you want to divide both sides by that variable, or multiply by the reciprocal (1/-E)

(VKC)/F=-EX

(1/-E)*(VKC/F)=(1/-E)*(-EX)

Multiply the numerators and denominators on the left hand, then cancel out the E's on the right hand.

VKC/-EF=X

- seversonLv 43 years ago
yet another technique from the attitude of producing applications: sum (2^n x^(2^n))/(a million + x^(2^n)) ... = x sum (2^n x^(2^n - a million))/(a million + x^(2^n)) ... = x sum D(log(a million + x^(2^n))) ... = x D(log(prod (a million + x^(2^n)))) the place D stands for differentiation with know to x. Use induction on okay to be sure that (a million + x)(a million + x^2)(a million + x^(2^2))...(a million + x^(2^ok)) = a million + x + x^2 + x^3 + ... + x^(2^(ok + a million) - a million) As ok (or n) has a tendency to infinity, we for this reason have prod (a million + x^(2^n)) = a million + x + x^2 + x^3 + .... = a million/(a million - x) for |x| < a million. for that reason, for |x| < a million, that's composable with log and differentiable, and so the unique sequence evaluates to x D(log(prod (a million + x^(2^n)))) ... = x D(log(a million/(a million - x))) ... = x(a million - x)/(a million - x)^2 ... = x/(a million - x)