Answer:
[tex]\mathsf e^{t-1}[/tex]
Step-by-step explanation:
Definition of an inverse function
[tex]\textsf {A\:function\:g\:is\:the\:inverse\:of\:function\:f\:if\:for}\:y=f\left(x\right),\:\:x=g\left(y\right)\:[/tex]
In order to find the inverse of p(t) = 1 + ln(t)...
- Set y = 1 + ln(t)
- Replace t with y
[tex]\textsf t=1+\ln \left(y\right)[/tex] - Solve for y
[tex]\textsf 1+\ln \left(y\right)=t[/tex] - Subtract 1 from both sides
[tex]\textsf\ln \left(y\right)=t-1[/tex][tex]\textsf {$\ln \left(y\right)=t-1$}[/tex] - Apply the log rule: [tex]\mathsf{If}\:\log _a\left(b\right)=c\:\mathsf{then}\:b=a^c[/tex]
[tex]\ln \left(y\right)=t-1\quad[/tex]
[tex]\implies y=e^{t-1}[/tex]
