Practice Problems for 1/30
- Find a formula for the inverse of the function \( f(x)=3x-2\).
Solve for \(x\) in terms of \(y\): \(y=3x-2 \implies y+2=3x\), so \(x=\frac{y+2}{3}\). Now, switch \(x\) and \(y\) to get the formula for the inverse: \(f^{-1}(x)=\frac{x+2}{3}\).
- Use the properties of logarithms to solve for x if
\[
\log(3x)=\log(x+2)-\log(4),
\]
where \( \log\) means logarithm base 10.
\(\begin{align*}
\log(3x)&= \log(x+2)-\log(4)\\
&= \log \left( \frac{x+2}{4}\right).
\end{align*}
\)
Since logarithmic functions are one-to-one, this implies \(3x=\frac{x+2}{4}\). Solving this simplier equation for \(x\) gives \(x=2/11\).
-
Simplify the following expressions.
- \(\log_3(27)\)
\(\log_3(27)=\log_3(3^3)=3\log_3(3)=3(1)=3.\)
- \(16^{\log_4(5)}\)
\(
\begin{align*}
16^{\log_4(5)}&=(4^2)^{\log_4(5)}\\
&=4^{2\log_4(5)}\\
&=4^{\log_4(5^2)}\\
&=4^{\log_4(25)}=25.
\end{align*}
\)
Or,
\(
\begin{align*}
16^{\log_4(5)}&=(4^2)^{\log_4(5)}\\
&=4^{2\log_4(5)}\\
&=(4^{\log_4(5)})^2\\
&=(5)^2=25.
\end{align*}
\)
- History of Logarithms.
Prosthaphaeresis is the method of multiplying large numbers together by using trigonometric identities which turn multiplication of cosines and sines into addition of cosines and sines, such as the identity
\[
\cos a \cos b=\frac{\cos (a+b)+\cos(a-b)}{2}.
\]
If one wished to multiply two numbers \(x\) and \(y\) one would first have to use trig tables to find numbers \(a\) and \(b\) such that \(x=\cos a\) and \(y=\cos b\). Then \(xy=\cos a \cos b\), so the above identity would be applied using trig tables to calculate \(\cos (a+b)\) and \(\cos(a-b)\).
This is a lot of work and was the main method of simplifying multiplication before logarithms were invented, and even then it was necessary to look up numbers in tables. The links below provide a brief description of prosthaphaeresis and the history of logarithms. They're worth a quick glance if you're curious about history or applications of formulas and properties presented in most every precalculus textbook.