Solve for the indicated variable.
-$A = \frac { t s } { 1 - ( 1 + z ) ^ { - B } }$, for B
A) $B = \frac { \ln A - \ln ( A - t s ) } { \ln ( 1 + z ) }$
B) $B = \ln \frac { t } { ( A - A s ) ( 1 + z ) }$
C) $B = \frac { \ln t } { \ln [ ( A - A s ) ( 1 - z ) ] }$
D) $B = \ln \frac { A } { ( A - t s ) ( 1 + z ) }$

Question

Quizplus · Accepted Answer

To solve for $B$, we start with the given equation $A = \frac { t s } { 1 - ( 1 + z ) ^ { - B } }$ and rearrange it to isolate $B$. First, multiply both sides by the denominator and then divide by $A$ to get $1 - (1+z)^{-B} = \frac{ts}{A}$. Rearrange to get $(1+z)^{-B} = 1 - \frac{ts}{A}$, which simplifies to $(1+z)^{-B} = \frac{A-ts}{A}$. Taking the natural logarithm of both sides gives $-B \ln(1+z) = \ln\left(\frac{A-ts}{A}\right)$, which can be rearranged to $B = \frac{\ln A - \ln(A-ts)}{\ln(1+z)}$, matching choice A.

Solve for the Indicated Variable A=ts1−(1+z)−BA = \frac { t s } { 1 - ( 1 + z ) ^ { - B } }A=1−(1+z)−Bts​

Solve for the Indicated Variable $A = \frac { t s } { 1 - ( 1 + z ) ^ { - B } }$