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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: bx = y.

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

Example: if b = 2 and x = 8, then log2(8) = 3 since 3 satisfies the equation 2x = 8.

Another example: log10(100) = 2 since 102 = 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 loge(y) is ln(y). The standard notation for log10(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: bxbw = bx+w

Suppose we have two fixed values y and z and suppose we know that logb(y) = x and logb(z) = w. This means that bx = y and bw = z. If we multiply y and z, we get yz = bxbw = bx+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,

logb(yz) = logb(y) + logb(z).

This is the product rule for logarithms. In words, logb(yz) = logb(y) + logb(z) means that the logarithm of a product is equal to the sum of the logarithms of the factors.

Let’s compute log7(7*49). We notice that log7(7) = 1 and log7(49) = 2.

Thus, log7(7*49) = log7(7) + log7(49) = 1+2 = 3.

Quotient Identity

This identity comes from the exponent rule for products: (bx)/(bw)=bx-w

Just as before, if logb(y) = x and logb(z) = w. This means that bx = y and bw = z. If we divide y and z, we get y/z = (bx)/(bw) = bx-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,

logb(y/z) = logb(y) - logb(z).

This is the quotient rule for logarithms. In words, logb(y/z) = logb(y) - logb(z) means that the logarithm of a quotient is equal to the difference of the logarithms of the factors.

Let’s compute log10(1/10,000). We notice that log10(1) = 0 and log10(10,000) = 4.

Thus, log10(1/10,000) = log10(1) - log10(10,000) = 0 - 4 = -4. You can check with a calculator that 10(-4) = 1/10,000.

Power Identity

Notice that, by = a implies that (by)x = byx = ax. This means that the exponent on b which causes ax is the same as x times the logarithm base b of a. This gives us the power identity for logarithmic functions is:

logb(ax) = x*logb(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: log2(165) = 5*log2(16) = 5log2(24) = 5*4*log2(2) = 5*4*1 = 20.

A couple important things to note about the power identity:

  1. The exponents which you “pull out” do not have to be whole numbers. For example: log15(225-4) = -4*log15(225) and log3(90.125) = 0.125log3(9).
  2. 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 logb( (w+y+z)a ) = a*logb(w+y+z)
    --However, if the fixed quantity is wa + ya, then logb(wa + ya) cannot be simplified.

Inverse functions

The power rule for the logarithm gives us that

logb(bx) = xlogb(b) = x.

This means that the inverse function to logb(y) is bx. On the other hand, logb(y) is the exponent we apply to b to get y. That means that

y = blogby

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 logb on one side, you can exponentiate the entire equation base b to “cancel” out that logarithm. Here’s what I mean:

Start with log13(x) = 25. Exponentiate both sides with respect to 13 to get: 13log13x = 1325. Canceling out the 13 and log13 we get: x = 1325.

Similarly, if we start with 8x = 64 and then take the logarithm of both sides we get log8(8x) = log8(64). Canceling out the log8 and the 8 on the right, we get x = log8(64) = 2.

Changing Bases

There is a straightforward equation for computing the logb(x) with respect to another basis, k:

logb(x) = (logk(x))/(logk(b)).

For example, log16(32) = log2(32)/log2(16) = 5/4.

Logarithmic Functions Practice Quiz

ln(y) = loge(y)

A. True
B. False
The correct answer here would be A.

ln(ex) = loge(ex) = x

A. True
B. False
The correct answer here would be A.

log10(100) = 3

A. True
B. False
The correct answer here would be B.

log3(27) = 3

A. True
B. False
The correct answer here would be A.

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

A. True
B. False
The correct answer here would be A.

log15(25/4) = log15(25) + log15(4).

A. True
B. False
The correct answer here would be B.

log5(25) = 2*log5(5)

A. True
B. False
The correct answer here would be A.

log5(125) = 5*log5(25)

A. True
B. False
The correct answer here would be B.

log6(1/6) = (log10(1/6))/(log10(6))

A. True
B. False
The correct answer here would be A.
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