91Ó°ÊÓ

If \(\$ 1\) is invested over a 10-year period, then the balance \(A,\) where \(t\) represents the time in years, is given by \(A=1+0.075[t]\) or \(A=e^{0.07 t}\) depending on whether the interest is simple interest at \(7 \frac{1}{2} \%\) or continuous compound interest at \(7 \%\) Graph each function on the same set of axes. Which grows at a greater rate? (Remember that \([t]]\) is the greatest integer function discussed in Section 1.6.)

Use the properties of logarithms to rewrite and simplify the logarithmic expression. $$\ln \left(5 e^{6}\right)$$

Use a graphing utility to construct a table of values for the function. Then sketch the graph of the function. $$f(x)=6^{-x}$$

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$4\left(3^{x}\right)=20$$

Evaluate the function at the indicated value of \(x\) without using a calculator. $$\begin{array}{cc}\text{Function} && \text {Value} \\ g(x)=\log _{a} x && x=a^{2} \end{array}$$

Use a graphing utility to construct a table of values for the function. Then sketch the graph of the function. $$f(x)=6^{x}$$

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$4 e^{x}=91$$

Evaluate the function at the indicated value of \(x\) without using a calculator. $$\begin{array}{cc}\text{Function} && \text {Value} \\ g(x)=\log _{b} x && x=b^{-3} \end{array}$$

Use the properties of logarithms to rewrite and simplify the logarithmic expression. $$\ln \frac{6}{e^{2}}$$

If \(\$ 1\) is invested over a 10-year period, then the balance \(A,\) where \(t\) represents the time in years, is given by \(A=1+0.06[t][\) or \(A=[1+(0.055 / 365)]^{[365 t]}\) depending on whether the interest is simple interest at \(6 \%\) or compound interest at \(5 \frac{1}{2} \%\) compounded daily. Use a graphing utility to graph each function in the same viewing window. Which grows at a greater rate?