# Logarithmic Functions

### Written by tutor Kira L.

A logarithmic function has three main components. The first component is the base, b; the second component is the fixed

value, y, which is what you input into the function; and the third component is the output of the logarithm function, x.

The output of the logarithm function is the answer to the following question: to what exponent must I raise the base, b,

in order to achieve the fixed value y? That is, the logarithm with base b of y is the solution to this equation: b^{x} = y.

The conventional notation is log_{b}(y) = x, which is read aloud as “log base b of y is equal to x.”

Example: if b = 2 and x = 8, then log_{2}(8) = 3 since 3 satisfies the equation 2^{x} = 8.

Another example: log_{10}(100) = 2 since 10^{2} = 100.

There the most commonly used bases for logarithms are base b=2, base b=10, and base b=e (where e is the constant approximately

equal to 2.718). The logarithm base e is commonly referred to as the natural logarithm and has many applications in pure

mathematics and calculus. The standard notation for log_{e}(y) is ln(y). The standard notation for log_{10}(y) is log(y).

If there is no base given, you assume it is base 10.

Now that we’ve gotten through the basics of what a logarithm is, let’s look at a few logarithmic identities.

## Product Identity

This identity comes from the exponent rule for products: b^{x}b^{w} = b^{x+w}

Suppose we have two fixed values y and z and suppose we know that log_{b}(y) = x and log_{b}(z) = w.

This means that b^{x} = y and b^{w} = z. If we multiply y and z, we get yz = b^{x}b^{w} = b^{x+w}. This means

that x+w is the exponent which we must raise b to in order to achieve the fixed value of yz. That is,

log_{b}(yz) = log_{b}(y) + log_{b}(z).

This is the product rule for logarithms. In words, log_{b}(yz) = log_{b}(y) + log_{b}(z) means that the logarithm of a product

is equal to the sum of the logarithms of the factors.

Let’s compute log_{7}(7*49). We notice that log_{7}(7) = 1 and log_{7}(49) = 2.

Thus, log_{7}(7*49) = log_{7}(7) + log_{7}(49) = 1+2 = 3.

## Quotient Identity

This identity comes from the exponent rule for products: (b^{x})/(b^{w})=b^{x-w}

Just as before, if log_{b}(y) = x and log_{b}(z) = w. This means that b^{x} = y and b^{w} = z. If we divide y and z,

we get y/z = (b^{x})/(b^{w}) = b^{x-w}.

This means that x-w is the exponent which we must raise b to in order to achieve the fixed value of y/z. That is,

log_{b}(y/z) = log_{b}(y) – log_{b}(z).

This is the quotient rule for logarithms. In words, log_{b}(y/z) = log_{b}(y) – log_{b}(z) means that the logarithm of a quotient is

equal to the difference of the logarithms of the factors.

Let’s compute log_{10}(1/10,000). We notice that log_{10}(1) = 0 and log_{10}(10,000) = 4.

Thus, log_{10}(1/10,000) = log_{10}(1) – log_{10}(10,000) = 0 – 4 = -4. You can check with a calculator that 10^{(-4)} = 1/10,000.

## Power Identity

Notice that, b^{y} = a implies that (b^{y})^{x} = b^{yx} = a^{x}. This means that the exponent on b which causes

a^{x} is the same as x times the logarithm base b of a. This gives us the power identity for logarithmic functions is:

log_{b}(a^{x}) = x*log_{b}(a).

When you are trying to simplify logarithmic functions, it is helpful to remember that when you see an exponent inside of a log, you can pull

that exponent out to the front of the function.

Example: log_{2}(16^{5}) = 5*log_{2}(16) = 5log_{2}(2^{4}) = 5*4*log_{2}(2) = 5*4*1 = 20.

A couple important things to note about the power identity:

- The exponents which you “pull out” do not have to be whole numbers. For example: log
_{15}(225^{-4}) = -4*log_{15}(225)

and log_{3}(90.125) = 0.125log_{3}(9). - You can only pull an exponent out if the entire fixed quantity has that exponent applied to it.
–For example, if the fixed quantity is (w+y+z)

^{a}, then log_{b}( (w+y+z)^{a}) = a*log_{b}(w+y+z)

–However, if the fixed quantity is w^{a}+ y^{a}, then log_{b}(w^{a}+ y^{a}) cannot be simplified.

## Inverse functions

The power rule for the logarithm gives us that

log_{b}(b^{x}) = xlog_{b}(b) = x.

This means that the inverse function to log_{b}(y) is b^{x}. On the other hand, log_{b}(y) is the exponent we apply to b to get y.

That means that

y = b^{logby}

This means that the inverse function to taking the x-th power of b is the logarithm function base b. When you are solving equations

and you’ve got a log_{b} on one side, you can exponentiate the entire equation base b to “cancel” out that logarithm. Here’s what I mean:

Start with log_{13}(x) = 25. Exponentiate both sides with respect to 13 to get: 13^{log13x} = 13^{25}.

Canceling out the 13 and log_{13} we get: x = 13^{25}.

Similarly, if we start with 8^{x} = 64 and then take the logarithm of both sides we get log_{8}(8^{x}) = log_{8}(64).

Canceling out the log8 and the 8 on the right, we get x = log_{8}(64) = 2.

## Changing Bases

There is a straightforward equation for computing the log_{b}(x) with respect to another basis, k:

log_{b}(x) = (log_{k}(x))/(log_{k}(b)).

For example, log_{16}(32) = log_{2}(32)/log_{2}(16) = 5/4.

## Logarithmic Functions Practice Quiz

ln(y) = log_{e}(y)

**A.**

True

**B.**

False

**A**.

ln(e^{x}) = log_{e}(e^{x}) = x

**A.**

True

**B.**

False

**A**.

log_{10}(100) = 3

**A.**

True

**B.**

False

**B**.

log_{3}(27) = 3

**A.**

True

**B.**

False

**A**.

log(1)+log(200) = log(200)

**A.**

True

**B.**

False

**A**.

log_{15}(25/4) = log_{15}(25) + log_{15}(4).

**A.**

True

**B.**

False

**B**.

log_{5}(25) = 2*log_{5}(5)

**A.**

True

**B.**

False

**A**.

log_{5}(125) = 5*log_{5}(25)

**A.**

True

**B.**

False

**B**.

log_{6}(1/6) = (log_{10}(1/6))/(log_{10}(6))

**A.**

True

**B.**

False

**A**.