16 Section 3.5

a. \begin{align*}
5(1.32)^t & = 12 \\
(1.32)^t & = \frac{12}{5} \\
\ln(1.32)^t & = \ln(\frac{12}{5}) \\
t\ln(1.32) & = \ln(\frac{12}{5}) \\
t & = \frac{\ln(\frac{12}{5})}{\ln(1.32)} \\
\end{align*}

b. \begin{align*}
3000(\frac{1}{2})^{\frac{t}{15}} & = 9000 \\
(\frac{1}{2})^{\frac{t}{15}} & = 3 \\
\ln((\frac{1}{2})^{\frac{t}{15}}) & = \ln(3) \\
\frac{t}{15}\ln(\frac{1}{2}) & = \ln(3) \\
t & = \frac{15\ln(3)}{\ln(\frac{1}{2}}) \\
\end{align*}

c. \begin{align*}
2^{x+5} & = 9 \\
\ln(2^{x+5}) & = \ln(9) \\
(x+5)\ln(2) & = \ln(9) \\
x+5 & = \frac{\ln(2)}{\ln(9)} \\
x & = -5 + \frac{\ln(2)}{\ln(9)} \\
\end{align*}

d. \begin{align*}
e^{3x} + 8 & = 12 \\
e^{3x} & = 4 \\
\ln(e^{3x}) & = \ln(4) \\
3x & = \ln(4) \\
x & = \frac{\ln(4)}{3}
\end{align*}

e. \begin{align*}
18e^{3t+1} & = 68 \\
e^{3t+1} & = \frac{34}{9} \\
\ln(e^{3t+1}) & = \ln(\frac{34}{9}) \\
3t+1 & = \ln(\frac{34}{9}) \\
3t & = -1 + \ln(\frac{34}{9}) \\
t &= \frac{-1 + \ln(\frac{34}{9})}{3} \\
\end{align*}

f. \begin{align*}
1.5(3^x) & = 2(4^x) \\
\frac{1.5}{2} & = \frac{4^x}{3^x} \\
\frac{ 3}{4} & = \frac{4^x}{3^x} \\
\ln(\frac{3}{4}) & = \ln(\frac{4^x}{3^x}) \\
ln(\frac{3}{4}) & = \ln(4^x) – \ln(3^x) \\
\ln(\frac{3}{4}) & = x\ln(4) – x\ln(3) \\
\ln(\frac{3}{4}) & = x(\ln(4) – \ln(3)) \\
x &= \frac{\ln(\frac{3}{4})}{\ln(4)-\ln(3)} \\
\end{align*}

a. \begin{align*}
20\ln(x-3) & = 40 \\
\ln(x-3) & = 2 \\
e^2 & = x-3 \\
x & = e^2 + 3 \\
\end{align*}

b. \begin{align*}
\log(2-x) – \log(2+x) & = 3 \\
\log(\frac{2-x}{2+x}) & = 3 \\
10^3 & = \frac{2-x}{2+x} \\
1000(2+x) & = 2-x \\
2000+1000x & = 2-x \\
1001x & -1998 \\
x & = \frac{-1998}{1001} \\
\end{align*}

c. \begin{align*}
\log(x-5) + \log(x+4) & = 1 \\
\log((x-5)(x-4) & = 1 \\
10^{1} & = (x-5)(x+4) \\
10 & = x^2 -x-20 \\
0 & = x^2 – x – 30 \\
0 & = (x-6)(x+5) \\
\end{align*}

x = 6 is the only solution since x = -5 is not in the domain.

a. [latex]y = 3\ln(x-4)[/latex]

• Vertical Stretch of 3, Up 4
• Graph
• VA: [latex]x = 4[/latex]
• Domain: [latex](4, \infty)[/latex]
• x-intercept: [latex](5,0)[/latex] and no y-intercept

b. [latex]y = 2-\ln(3-x)[/latex]

• right 3, reflection across y-axis, up 2
• Graph
• VA: x = 3
• Domain: [latex](-\infty, 3)[/latex]
• x-intercept: [latex](3-e^2, 0)[/latex], y-intercept: [latex](0, 2-\ln(3))[/latex]

a. We will find a function of the form [latex]f(t) = ab^t[/latex]. We know that [latex]a = 3000[/latex]. So now we solve for [latex]b[/latex]. So, [latex]1500 = 3000(b)^{5730}[/latex]. Then solving for [latex]b[/latex] we get [latex]b =  \frac{1}{2}^{\frac{1}{5730}}[/latex]. Thus, [latex]f(t) = 3000(\frac{1}{2}^{\frac{1}{5730}})^t[/latex]

We will find a model of the form [latex]f(t) = ab^t[/latex]. We know that since [latex]r = 3[/latex]%, then [latex]b = 1.03[/latex]. Then to solve for the doubling time [latex]t[/latex], we set the equation equal to [latex]2a[/latex] to get [latex]2a = a(1.03)^t[/latex], then dividing by [latex]a[/latex] we have [latex]2 = (1.03)^t[/latex]. Next, taking the natural log of both sides we get [latex]\ln(2) = \ln(1.03)^t[/latex]. Then solving for [latex]t[/latex] we get [latex]t = \frac{\ln(1.03)}{\ln(2)} \approx 0.0426[/latex] years.

a. We know that [latex]a = 15[/latex]. Plugging in the point [latex](25, 150)[/latex] we have [latex]150 = 15e^{k\cdot 25}[/latex]. Then we divide by 15 on both sides to get [latex]10 = e^{k \cdot 25}[/latex]. Then taking the natural log of both sides we get [latex]\ln(10) = \ln(e^{k \cdot 25})[/latex]. Then solving for [latex]k[/latex] we get [latex]\frac{\ln(10)}{25}[/latex]. So, [latex]F(t) = 15e^{k \cdot \frac{\ln(10)}{25}}[/latex].

b. Setting our equation equal to 175 we get  [latex]175 = 15e^{t \cdot \frac{\ln(10)}{25}}[/latex]. Then we divide by 15 on both sides and take the natural log of both sides to get [latex]\ln(\frac{35}{3}) = \ln(e^{ t \cdot \frac{\ln(10)}{25}})[/latex]. Then simplifying we get [latex]\ln(\frac{35}{3}) = t  \cdot \frac{\ln(10)}{25}[/latex]. So, [latex]t \approx \frac{\ln(\frac{35}{3})}{\frac{\ln(10)}{25}} \approx 26.67[/latex] minutes.