how do I rearrange the following eqation so that it starts Bt = ? K=1/t * ln(Bo/Bt)?

Answer 1

K = 1/t * ln(Bo/Bt)

multiply both sides by t

Kt = ln(Bo/Bt)

exponentialise both sides

e^(Kt) = Bo/Bt

multiply both sides by Bt and divide both by e^(Kt)

Bt = Bo/(e^(Kt))

Hope the explanation helps...

Answer 2

K=1/t*ln(Bo/Bt)

is the same as

K*t= ln(Bo/Bt)

Now you need to perform the inverse function of ln to remove the ln from the equation i.e. both sides become powers to which e is raised

e to the power of K*t = e to the power of ln(Bo/Bt)

e to the power of ln are inverse functions so they cancel leaving

e to the K*t = Bo/Bt

So Bo/e to the K*t = Bt

P.S. I hope the original equation reads 1/t multiplied by ln(Bo/Bt) and not 1 divided by t*ln(B0/Bt)

Answer 3

This seems to be ambiguous;

Is the equation Bt=1/(t*ln(Bo/Bt) or
Bt=(1/t)*{ln(Bo/Bt)} = {ln(Bo/Bt)}/t
These 2 are different.
The answers so far have assumed the first.

Answer 4

K = 1/t * ln(Bo/Bt)

=> t/K = ln(Bo/Bt)

=> e^(t/K) = Bo/Bt

=> Bt = Bo / (e^(t/K) )

Answer 5

K=1/t * ln(Bo/Bt)

K/(1/t) =ln(Bo/Bt)
K* t=ln(Bo/Bt)
e^(K*t) = Bo/Bt
e^(K*t)/Bo =1/Bt

Bt = Bo/(e^(K*t))

Answer 6

K = 1/t*ln(Bo/Bt)

(a) Kt = ln (Bo/Bt)

(b) e^(Kt) = Bo/Bt

(c) Bt* e^(Kt) = Bo

(d) Bt = Bo/e^(Kt)

Answer 7

k = (1 / t) ln (Bo / Bt)
t k = ln (Bo / Bt)
e^(t k) = Bo / Bt
Bt = Bo / ( e^(tk) )

Answer 8

K=1/t * ln(Bo/Bt)
Kt = ln(Bo/Bt)
Bo/Bt = e^(Kt)
Bo/[e^(Kt)] = Bt
Bt = Bo[e^(-Kt)]