Solve for the indicated variable.
-$A = \frac { W B } { 1 - ( 1 + z ) ^ { - s ^ { \prime } } }$, for $\mathrm { s }$
A) $s = \frac { \ln A - \ln ( A - W B ) } { \ln ( 1 + z ) }$

B) $s = \ln \frac { W } { ( A - A B ) ( 1 + z ) }$

C) $s = \frac { \ln W } { \ln [ ( A - A B ) ( 1 - z ) ] }$

D) $s = \ln \frac { A } { ( A - W B ) ( 1 + z ) }$

Question

Solve for the indicated variable.
-$A = \frac { W B } { 1 - ( 1 + z ) ^ { - s ^ { \prime } } }$, for $\mathrm { s }$
A) $s = \frac { \ln A - \ln ( A - W B ) } { \ln ( 1 + z ) }$

B) $s = \ln \frac { W } { ( A - A B ) ( 1 + z ) }$

C) $s = \frac { \ln W } { \ln [ ( A - A B ) ( 1 - z ) ] }$

D) $s = \ln \frac { A } { ( A - W B ) ( 1 + z ) }$

Quizplus · Accepted Answer

To solve for $s'$, first isolate the term containing $s'$ by rearranging the equation: $A = \frac{WB}{1 - (1 + z)^{-s'}}$ becomes $1 - (1 + z)^{-s'} = \frac{WB}{A}$, which simplifies to $(1 + z)^{-s'} = 1 - \frac{WB}{A}$. Taking the natural logarithm of both sides and solving for $s'$ yields $s' = \frac{\ln A - \ln(A - WB)}{\ln(1 + z)}$, which matches choice A.

Solve for the Indicated Variable A=WB1−(1+z)−s′A = \frac { W B } { 1 - ( 1 + z ) ^ { - s ^ { \prime } } }A=1−(1+z)−s′WB​

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