1. \(8^{1/3} = 2\) is equivalent to \(\log_8 2 = \frac{1}{3}\).

2. \(\log_2 \frac{1}{32} = -5\) is equivalent to \(2^{-5} = \frac{1}{32}\).

3. \(\log_2 x = 3\) is equivalent to \(x = 2^3 = 8\).

4. \(\log_2 4 = x\) is equivalent to \(2^x = 4\) and we see that \(x=2\) since \(2^2 = 4\).

5. Using the product and power properties \begin{equation*} \log\left( x^3 y^5 \right) = 3\log(x) + 5\log(y). \end{equation*}

6. Using the product, power, and quotient properties \begin{equation*} \log\left( \frac{x^{15} y^{20}}{z^4} \right) = 15\log(x) + 20\log(y) - 4\log(z). \end{equation*}

7. Using the power and product properties \begin{equation*} \ln(8) + 5 \ln(x) + 15\ln(x^2+8) = \ln\left( 8 x^5 \left( x^2 + 8 \right)^{15} \right) \end{equation*}

To isolate the variable \(x\text{,}\) we should take the logarithm of both sides. For convenience, let's choose to use the natural logarithm.

Applying logarithm property (5), we find

Solving algebraically for \(x\) yields

NOTE: Our choice to use the natural logarithm was arbitrary. We could have chosen any base for our logarithm to solve this equation.

Since the rate of growth of the population is proportional to the size of the population, we should use an exponential model for this problem. That is, we want

where \(t\) is the number of years after 1970, \(P(t)\) is the population (in millions) of the United States at time \(t\text{,}\) \(P_{0}\) is the population (in millions) of the United States in 1970 (i.e. \(t=0\)) and \(r\) is the continuous rate of growth of the population. To determine when the population will reach 350 million, we must solve the equation

To solve for \(t\text{,}\) we need to first solve for the exponential expression by itself and then use logarithms. Dividing both sides of the equation by 205.1 gives

Taking the natural logarithm of both sides gives

Applying logarithm property (6), we find

Finally, solving algebraically for \(t\) gives

Thus, we expect the population of the United States to reach 350 million in 2021 (approximately 51 years after 1970).

1. \(t = e^4\)

2. \(t+3 = e^4\) so \(t = e^4 - 3\)

3. \(t+3 = 4\) so \(t=1\)

4. Using the product property for logarithms, \(\ln(t(t+3)) = \ln(4)\) so \(t^2 + 3t = 4\). This is quadratic so we rearrange to get \(t^2 + 3t - 4 = 0\) which factors to \((t+4)(t-1) = 0\) so the solutions are \(t=1\) and \(t=-4\). Substituting back to the original equation we see that \(\ln(-4+3)\) and \(\ln(-4)\) do not exist so the only solution is \(t=1\).

5. \(t=\ln(4)\)

6. \(t= \ln(4) = 3\)

7. \(t+3 = \ln(2)\) so \(t = \ln(2) = 3\)

8. For the final problem we first divide by \(2\) and take the natural logarithm of both sides. \begin{equation*} e^{3t+1} = \frac{3}{2} e^{t-1} \implies 3t+1 = \ln(3/2) + t-1. \end{equation*} Therefore, \(2t = \ln(3/2) - 2\) so \(t=\frac{1}{2} \left( \ln(3/2) - 2 \right)\).

1. We expect 1 solution since the exponential function \(7^x\) is always increasing.

2. \(\log_7 (24) = x\).

3. \(\log (7^x) = \log(24)\) so \(x\log(7) = \log(24)\) and therefore \(x = \frac{\log(24)}{\log(7)}\)

4. \(\ln (7^x) = \ln(24)\) so \(x\ln(7) = \ln(24)\) and therefore \(x = \frac{\ln(24)}{\ln(7)}\)

5. \(\log_7(x) = \frac{\ln(x)}{\ln(7)} = \frac{\log(x)}{\log(7)}\).

1. If the population doubles every three hours then the exponential function describing the growth is \(P(t) = P_0 2^{t/3}\). We want to find the continuous growth rate so we want to write the function as \(P(t) = P_0 e^{kt}\). Therefore we need to solve the equation \(2^{t/3} = e^{kt}\) for \(k\). Taking the natural logarithm of both sides gives \(\ln \left( 2^{t/3} \right) = kt\), and using the rules of exponents we see that \(\frac{t}{3} \ln(2) = kt\). Subtracting \(kt\) from both sides and factoring the \(t\) gives \(t\left( \frac{\ln(2)}{3} - k \right) = 0\) so we immediately see that \(k = \ln(2)/3\).

2. We need to solve the equation \(65+186e^{-0.06t} = 120\). Subtracting 65 and dividing by 186 gives \(e^{-0.06t} = \frac{55}{186}\). Taking the natural logarithm of both sides and then dividing by -0.06 gives \(t = -\frac{1}{0.06} \ln \left( \frac{55}{186} \right)\).

3. We need to solve the equation \(80\left( 1-e^{-0.2t} \right) = 75\). Dividing by 80 gives \(1-e^{-0.2t} = \frac{15}{16}\). Hence, \(e^{-0.2t} = \frac{1}{16}\) and taking the natural logarithm of both sides gives \(-0.2t = \ln(1/16)\). Therefore, \(t = -\frac{1}{0.2} \ln\left( \frac{1}{16} \right)\).