Hi, can anyone help me rearrange this equation for B please? A = B * exp(15(B-1)) Thanks?

Answer 1

A = B e^(15B) e^(-15)
B = (e^15 ) A e^(-15B)
= (e^15 ) A [ 1 -15B + (15B)^2/2! - .......]
= A e^15 - 15 A B e^15 + .......
= A e^15 - 15 A [ A e^15 - 15 A B e^15 + ....... ] e^15 + .......
= A e^15 - 15 A^2 (e^15)^2 + .......

The functin on the right is not a one-to-one function, and hence you cannot expect to have a inverse. You must restrict the domain (for example, B > -1/15 or B < -1/15) to invert the formula.

The answer is also not that straight forward. What I showed above is a way to express B as a power series in A, I calculated up to A^2 term.

Answer 2

a + 15 = b * 15b... is that what you wanted?

or you can get

a + 15 = 15b squared

Answer 3

if you mean

A=B* e^(15(B-1))
A/B= e^(15(B-1))

ln(A/B) = 15 (B -1)

ln(A/B)+15=B

A/B +e^15 =e^B

A + B(e^15)=B *e^B

A=B(e^B-e^15)

im afraid you might be stuck here
although you could use mclaurin series to approximate answer