91Ó°ÊÓ

Use the change-of-base theorem to find an approximation to four decimal places for each logarithm. $$\log _{0.32} 5$$

Given the approximations \(\log _{10} 2=0.3010\) and \(\log _{10} 3=0.4771,\) find logarithm without using a calculator. $$\log _{10} \frac{9}{4}$$

Answer each of the following. Suppose \(f(r)\) is the volume (in cubic inches) of a sphere of radius \(r\) inches. What does \(f^{-1}(5)\) represent?

Given the approximations \(\log _{10} 2=0.3010\) and \(\log _{10} 3=0.4771,\) find logarithm without using a calculator. $$\log _{10} \frac{20}{27}$$

Given the approximations \(\log _{10} 2=0.3010\) and \(\log _{10} 3=0.4771,\) find logarithm without using a calculator. $$\log _{10} \sqrt{30}$$

Let \(u=\ln a\) and \(v=\ln b .\) Write each expression in terms of \(u\) and \(v\) without using the In function. $$\ln \left(b^{4} \sqrt{a}\right)$$

Answer each of the following. For a one-to-one function \(f,\) find \(\left(f^{-1} \circ f\right)(2),\) where \(f(2)=3\).

Let \(u=\ln a\) and \(v=\ln b .\) Write each expression in terms of \(u\) and \(v\) without using the In function. $$\ln \frac{a^{3}}{b^{2}}$$

Given the approximations \(\log _{10} 2=0.3010\) and \(\log _{10} 3=0.4771,\) find logarithm without using a calculator. $$\log _{10} 36^{1 / 3}$$

Use a graphing calculator to graph each function defined as follows, using the given viewing window. Use the graph to decide which functions are one-to-one. If a function is one-to-one, give the equation of its inverse. $$\begin{aligned}&f(x)=6 x^{3}+11 x^{2}-6;\\\&[-3,2] \text { by }[-10,10]\end{aligned}$$