Answer
a.
In month 4, $15%$ of those who quit and used telephone counseling will still be smoke-free.
b.
In month 16, $5%$ of those who quit and used telephone counseling will still be smoke-free.
Work Step by Step
a.
Consider the model $P\left( t \right)=21.4{{\left( 0.914 \right)}^{t}}$
Here, $P\left( t \right)$ is the percentage of smokers who receive telephone counseling and successfully quit smoking for $t$ months.
Replace $P\left( t \right)$ with 15 and solve for $t$ to calculate that month in which $15%$ percent of those who quit and used telephone counseling will still be smoke-free.
$\begin{align}
& P\left( t \right)=21.4{{\left( 0.914 \right)}^{t}} \\
& 15=21.4{{\left( 0.914 \right)}^{t}}
\end{align}$
Divide both the sides by $21.4$:
$\begin{align}
& \frac{15}{21.4}=\frac{21.4{{\left( 0.914 \right)}^{t}}}{21.4} \\
& 0.700={{\left( 0.914 \right)}^{t}}
\end{align}$
Take the common logarithm $\log $on both the sides:
$\log 0.700=\log {{\left( 0.914 \right)}^{t}}$
Apply the power rule of logarithms $\log {{m}^{n}}=n\log m$ in $\log {{\left( 0.914 \right)}^{t}}$:
$\begin{align}
& \log 0.700=\log {{\left( 0.914 \right)}^{t}} \\
& \log 0.700=t\log 0.914 \\
\end{align}$
Divide by $\log 0.914$ on both the sides;
$\begin{align}
& \frac{\log 0.700}{\log 0.914}=\frac{t\log 0.914}{\log 0.914} \\
& \frac{\log 0.700}{\log 0.914}=t
\end{align}$
Therefore,
$\begin{align}
& \frac{\log 0.700}{\log 0.914}=t \\
& \frac{-0.154}{-0.039}\approx t \\
& 3.94\approx t
\end{align}$
b.
Consider the model $P\left( t \right)=21.4{{\left( 0.914 \right)}^{t}}$
Here, $P\left( t \right)$ is the percentage of smokers who receive telephone counseling and successfully quit smoking for $t$ months.
Replace $P\left( t \right)$ with 5 and solve for $t$ to calculate that month in which $5\%$ of those who quit and used telephone counseling will still be smoke-free.
$\begin{align}
& P\left( t \right)=21.4{{\left( 0.914 \right)}^{t}} \\
& 5=21.4{{\left( 0.914 \right)}^{t}}
\end{align}$
Divide both the sides by $21.4$:
$\begin{align}
& \frac{5}{21.4}=\frac{21.4{{\left( 0.914 \right)}^{t}}}{21.4} \\
& 0.233={{\left( 0.914 \right)}^{t}}
\end{align}$
Take the common logarithm on both the sides:
$\log 0.233=\log {{\left( 0.914 \right)}^{t}}$
Apply the power rule of logarithms $\log {{m}^{n}}=n\log m$ in $\log {{\left( 0.914 \right)}^{t}}$:
$\begin{align}
& \log 0.233=\log {{\left( 0.914 \right)}^{t}} \\
& \log 0.233=t\log 0.914 \\
\end{align}$
Divide by $\log 0.914$ on both the sides;
$\begin{align}
& \frac{\log 0.233}{\log 0.914}=\frac{t\log 0.914}{\log 0.914} \\
& \frac{\log 0.233}{\log 0.914}=t
\end{align}$
Therefore,
$\begin{align}
& \frac{\log 0.700}{\log 0.914}=t \\
& \frac{-0.632}{-0.039}\approx t \\
& 16.2\approx t
\end{align}$
