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

Question

Quizplus · Accepted Answer

To solve for $t$, we start with the given equation: $$A = \frac { B s } { 1 - ( 1 + z ) ^ { - t ^ { \prime } } }$$First, we isolate the term containing $t$:$$A = \frac{Bs}{1 - (1+z)^{-t}}$$$$A(1 - (1+z)^{-t}) = Bs$$$$A - A(1+z)^{-t} = Bs$$$$A - Bs = A(1+z)^{-t}$$$$(1+z)^{-t} = \frac{A - Bs}{A}$$Taking the natural logarithm on both sides:$$-t \ln(1+z) = \ln\left(\frac{A - Bs}{A}\right)$$$$t = \frac{\ln\left(\frac{A - Bs}{A}\right)}{-\ln(1+z)}$$$$t = \frac{\ln(A) - \ln(A - Bs)}{\ln(1+z)}$$This matches choice A.

Solve for the Indicated Variable A=Bs1−(1+z)−t′ for tA = \frac { B s } { 1 - ( 1 + z ) ^ { - t ^ { \prime } } } \text { for } tA=1−(1+z)−t′Bs​ for t

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