Answer:
C. P–1(t) = log5 (StartFraction t Over 19,300 EndFraction)
Step-by-step explanation:
The inverse function of P(t) is [tex]P^{-1} (t)=log_{5}( \frac{t}{19300})[/tex]
The inverse function is a function obtained by reversing the given function. An inverse is a function that serves to “undo” another function. The domain of the original function becomes the range of inverse function and the range of the given function becomes the domain of the inverse function.
For the given situation,
The function P(t) = 19,300(5)^t, where t is the elapsed time in hours.
Steps to find inverse function:
Step 1: Replace P(t) as y
[tex]P(t) = 19,300(5)^t[/tex]
⇒ [tex]y= 19,300(5)^{t}[/tex]
Step 2: interchange t and y
⇒ [tex]t = 19,300(5)^y[/tex]
Step 3: Now solve for y
⇒ [tex]\frac{t}{19300} =5^y[/tex]
By law of logarithm,
[tex]a^{x} = y[/tex]
⇒ [tex]log_{a} y=x[/tex]
Thus the equation becomes,
⇒ [tex]log_{5} \frac{t}{19300}=y[/tex]
Step 4: Now replace y as [tex]P^{-1}( t)[/tex]
⇒ [tex]P^{-1} (t)=log_{5}( \frac{t}{19300})[/tex]
Hence we can conclude that the inverse function of P(t) is [tex]P^{-1} (t)=log_{5}( \frac{t}{19300})[/tex].
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