91Ó°ÊÓ

If $$\$ 1$$ is invested in an account over a 10-year period, the amount in the account, where \(t\) represents the time in years, is given by \(A=1+0.075 \llbracket t \rrbracket\) or \(A=e^{0.07 t}\) depending on whether the account pays 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 higher rate? (Remember that \(\llbracket t \rrbracket\) is the greatest integer function discussed in Section 1.6.)

Use a graphing utility to graph the exponential function. $$y=3^{x-2}+1$$

Find the exact value of the logarithmic expression without using a calculator. (If this is not possible, state the reason.) $$\log _{2} \sqrt[4]{8}$$

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$e^{x^{2}-3}=e^{x-2}$$

Use a calculator to evaluate \(f(x)=\log x\) at the indicated value of \(x\). Round your result to three decimal places. $$x=12.5$$

Use a calculator to evaluate \(f(x)=\log x\) at the indicated value of \(x\). Round your result to three decimal places. $$x=96.75$$

Solve the exponential equation algebraically. Approximate the result to three decimal places. $$e^{-x^{2}}=e^{x^{2}-2 x}$$

Use a graphing utility to graph the exponential function. $$y=4^{x+1}-2$$

COMPARING MODELS If $$\$ 1$$ is invested in an account over a 10 -year period, the amount in the account, where \(t\) represents the time in years, is given by \(A=1+0.06 \llbracket t \rrbracket\) or \(A=[1+(0.055 / 365)]^{[365 t]}\) depending on whether the account pays 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 higher rate?

Find the exact value of the logarithmic expression without using a calculator. (If this is not possible, state the reason.) $$\log _{6} \sqrt[3]{6}$$