Solve for the indicated variable.
-$\mathrm { V } - \mathrm { V } _ { 0 } = \left( \mathrm { V } _ { 1 } - \mathrm { V } _ { 0 } \right) 10 ^ { - \mathrm { kt } }$, for $\mathrm { t }$
A) $t = - \frac { 1 } { k } \log \frac { V } { V _ { 1 } }$
B) $t = - \frac { 1 } { k } \log \left( V - V _ { 1 } \right)$
C) $t = - \frac { V - V _ { 0 } } { k \left( V _ { 1 } - V _ { 0 } \right) }$
D) $t = - \frac { 1 } { k } \log \frac { V - V _ { 0 } } { V _ { 1 } - V _ { 0 } }$

Question

Quizplus · Accepted Answer

To solve for $t$, first isolate the exponential term: $10^{-kt} = \frac{V - V_0}{V_1 - V_0}$. Then, apply the logarithm to both sides to solve for $t$, resulting in $t = -\frac{1}{k} \log \frac{V - V_0}{V_1 - V_0}$.

Solve for the Indicated Variable V−V0=(V1−V0)10−kt\mathrm { V } - \mathrm { V } _ { 0 } = \left( \mathrm { V } _ { 1 } - \mathrm { V } _ { 0 } \right) 10 ^ { - \mathrm { kt } }V−V0​=(V1​−V0​)10−kt

Solve for the Indicated Variable $\mathrm { V } - \mathrm { V } _ { 0 } = \left( \mathrm { V } _ { 1 } - \mathrm { V } _ { 0 } \right) 10 ^ { - \mathrm { kt } }$