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Solve for the Indicated Variable P=110e6t, for tP = 110 e ^ { 6 t } , \text { for } t

Question 19

Multiple Choice

Solve for the indicated variable.
- P=110e6t, for tP = 110 e ^ { 6 t } , \text { for } t


A) t=lnP+ln1106\mathrm { t } = \frac { \ln \mathrm { P } + \ln 110 } { 6 }
B) t=ln(P110) 6t = \frac { \ln ( P - 110 ) } { 6 }
C) t=P110e6t = \frac { P } { 110 e ^ { 6 } }
D) t=lnPln1106t = \frac { \ln P - \ln 110 } { 6 }

Correct Answer:

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