Recall that exponential functions are increasing functions when the base is greater than one and decreasing functions when the base is between zero and one. Thus, exponential functions are one-to-one functions. Recall also that the inverse of a one-to-one function is, itself, a function.

Recreate your sketch of the graph of \(y = 2^x\) on a square sheet of paper. Sketch the graph of the inverse function by reversing the ordered pairs of points lying on the graph of \(y = 2^x\). You can also see what the graph of the inverse will look like by holding the sheet of graph paper by its lower left and upper right corners and rotating \(180^\circ\) about the line \(y = x\) so that you are looking at the back of the graph paper. Hold the paper up to a light source until you can see the graph through the paper. What you see will be the graph of the inverse function.

See SolutionRecreate your sketch of the graph of \(y = \left( \frac{1}{2} \right)^x\) on a square sheet of paper. Sketch the graph of the inverse function.

See SolutionRecall the method for finding the equation of an inverse function. First, replace the \(f ( x )\) with \(y\). Then exchange each \(x\) with \(y\) and visa versa. Next, solve the new equation for \(y\). Then replace that \(y\) with \(f^{-1}( x )\).

Let us apply that method to the exponential function \(f ( x ) = b^x\).

- Step 1: \(y = b^x\)
- Step 2: \(x = b^y\)
- Step 3: Now we must solve this equation for \(y\)!! But how can we do this?

How can we solve step 2 for \(y\) ? We want to find the value of \(y\) which, when it is the exponent of \(b\), yields the value \(x\). Such an exponent is called a **logarithm**. In this case, \(y\) would be the logarithm base \(b\) of \(x\). That is, \(y = \log_b ( x )\).

Therefore,

- Step 3: \(y = \log_b ( x )\)
- Step 4: \(f^{-1}( x ) = \log_b ( x )\)

**The inverse of an exponential function is a logarithmic function.**

This implies that the equation \(x = b^y\) is equivalent to the equation \(y = \log_b ( x )\).

This is a useful equivalence--**remember it!**

For example, use the equivalence to find \(\log_3(9)\) by first setting \(y=\log_3(9)\) then writing the equivalent form \(3^y=9\). Since \(9=3^2\) this gives us \(3^y=3^2\) and since exponential functions are one-to-one functions \(3^y=3^2\) if and only if \(y=2\). So we have our solution: \(\log_3(9)=2\).

Find the following logarithms:

- \(\log_5 ( 25 )\)
- \(\log_5 ( 5 )\)
- \(\log_5 ( 1 / 5 )\)

Sketch the graphs of the following logarithmic functions:

- \(f ( x ) = \log_2 ( x )\)
- \(g ( x ) = \log_{0.5} ( x )\)

Logarithmic functions are one of a class of functions which are sometimes written without function parentheses.

If the *argument* of the function, which is what usually goes between the pair of function parentheses, consists of just a single term then we may omit the function parentheses. Later, we will see that the trigonometric functions also have this feature.

Thus we can write \(\log_b(x)\) as \(\log_bx\) if we wish, or \(\log_23x^2y\) instead of \(\log_2(3x^2y)\). But we cannot write \(\log_b(2x+3y)\) without function parentheses. If we wrote \(\log_b2x+3y\) this would mean the same thing as \(3y+\log_b2x\).

There are certain rules of logarithms which follow from the rules of exponents and the exponential-logarithmic equivalence mentioned above.

Recall, specifically, that \(a^m a^n = a^{m + n}\) and \(( a^m )^n = a^{m n}\).

Now, let \( m = \log_b p\) and \(n = \log_b q\). Then \(p = b^m\) and \(q = b^n\) are the equivalent exponential equations. Thus, \(pq = b^mb^n = b^{m + n} = b^{( \log_bp + \log_b q )}\). So \(pq = b^{( \log_bp + \log_b q )}\). Using the exponential-logarithmic equivalence to re-write this exponential statement as a logarithmic statement, we get the result \(\log_bp +\log_bq = \log_bpq\). This is usually written in the other order as a Rule of Logarithms:

There are several others, the most common being

\[\log_b\left(\frac{p}{q}\right)=\log_bp-\log_bq\]

\[\log_bp^r = r \log_b p \]

\[ \log_b1=0\]

\[\log_bb=1\]

Let \(m = \log_b p^r \). Then \(b^m = p^r\), so \(b^{m / r} = p\). So \(\log_bp = \frac{m}{r}\). Thus, \(m = r \log_bp\).

Therefore \(\log_b p^r = r \log_bp\).

What logarithm rule derives from the exponential-logarithmic equivalence and the fact that \(b_0 = 1\)?

What rule derives from the fact that \(b^1 = b\)?

See SolutionGiven that \(\log_b2 = 1.7095\) and \(\log_b3 = 2.7095\), use the rules of logarithms to find the following:

- \(\log_b6\)
- \(\log_b72\)
- \(\log_b1.5\)
- Find the value of \(b\).

Use various rules of logarithms to rewrite the following in terms of \(\log_7x, \log_7y\), and \(\log_7z\):

\(\log_7\left(\dfrac{x^2\sqrt{z}}{y^3}\right)\)

See SolutionUse various rules of logarithms to rewrite the following as a single logarithm:

\(\log_3x + 2 \log_3y - \frac{1}{2}\log_3z\)

See SolutionThere are two standard bases for logarithms, namely base \(10\), called the **common base**, and base \(e\) called the **natural base**.

When no base is specified, as in \(\log x\), the common base \(10\) is assumed.

The natural base \(e\) is specified by writing \(\ln x\) in place of \(\log_e x\).

Find the following without using a calculator:

- \(\log0.01\)
- \(\ln e^3\)