Logarithmic Identities - Logarithms - High School Algebra II Unlocked (2016)

High School Algebra II Unlocked (2016)

Chapter 5. Logarithms

Lesson 5.3. Logarithmic Identities

Logarithmic identities are true equations that illustrate patterns in relationships between logarithms or parts of logarithms. We have already explored a number of logarithmic identities in this chapter: logbb = 1, logb1 = 0, logbbx = x, logb(1/bx) = −x, and blogbx = x.

There are also logarithmic properties related to mathematical operations. When you take a logarithm of a product, a quotient, a power, or a root, an interesting relationship emerges between the components and/or logarithms of the components.

Logarithmic Identities with Operations

For any positive real number b, where b ≠ 1, any positive real numbers x and y, and any real number n:

Product property: logbxy = logbx + logby

Quotient property: logb(x/y) = logbx − logby

Power property: logbxn = nlogbx

Root property:

Using x = 8, y = 4, b = 2, and n = 3, demonstrate that each of the four logarithmic identities shown above holds true.

Product property: logbxy = logbx + logby

log2(8 ⋅ 4) = log28 + log24

Substitute 8 for x, 4 for y, and 2 for b.

log232 = log28 + log24

Multiply 8 ⋅ 4.

5 = 3 + 2

Evaluate each of the three logarithms.

5 = 5

Quotient property: logb(x/y) = logbx − logby

log2(8/4) = log28 − log24

Substitute 8 for x, 4 for y, and 2 for b.

log22 = log28 − log24

Simplify 8/4.

1 = 3 − 2

Evaluate each of the three logarithms.

1 = 1

Power property: logbxn = nlogbx

log283 = 3log28

Substitute 8 for x, 2 for b, and 3 for n.

log2512 = 3log28

Evaluate 83.

9 = 3 · 3

Evaluate each of the two logarithms.

9 = 9

For simpler arithmetic,
we could have
instead showed
that logbyn = nlogby:
log243 = 3log24. This
simplifies as log264 =
3log24, or 6 = 3 · 2.

Root property:

Substitute 8 for x, 2 for b, and 3 for n.

Evaluate .

1 = 3/3

Evaluate each of the two logarithms.

1 = 1

Notice that the root
property is actually just
the power property
with a fractional value
of n. The cube root is
the same as raising to
a power of 1/3,
so the expression
log2 is equivalent
to log281/3. Using the
power property, we
can rewrite that as
1/3 log28, which is
the same as .

Prove the quotient property using other logarithmic identities.

We must prove that logb(x/y) = logbx − logby. The fraction x/y is the same as the product x ⋅ 1/y, so we can use the product property of logarithms.

logb(x/y)

= logb(x ⋅ 1/y)

Rewrite x/y as x ⋅ 1/y.

= logbxy−1

Rewrite 1/y as y−1.

= logbx + logby−1

Use the logarithm product property.

= logbx + (−1)logby

Use the logarithm power property.

= logbx − logby

Addition of a negative is a subtraction.

Another important logarithmic identity involves changing the base.

Change of base property: logbx =

Use the change of base property to rewrite log2781 using a base of 3, then simplify.

log2781

=

Use the change of base property.

= 4/3

Evaluate each logarithm.

Let’s use the definition of logarithm to check our answer.

274/3 ≟ 81

Rewrite the equation log2781 = 4/3 as an exponential relationship.

≟ 81

Rewrite 274/3 as (271/3)4, or .

34 ≟ 81

Evaluate .

81 = 81

Evaluate 34.

Try applying the
change of base
property to the
logarithms in
Example 4.

If possible, choose a base that allows you to directly evaluate each of the logarithms in the fraction, as in Example 12. However, even when this is not possible, the change of base property is still helpful for evaluating logarithms using technology. Any logarithm can be rewritten as a ratio of common logarithms or as a ratio of natural logarithms, each of which can be easily calculated on a calculator, as long as it has a “log” key or an “ln” key.

What is the approximate value of log316?

The numbers 3 and 16 do not share any common factors besides 1, so there is not a convenient new base to use to solve this by hand. Rewrite the given logarithm using the change of base property, as a ratio of common logarithms.

log316 =

Use your calculator to find the values of log16 and log3, and to perform the division. log316 ≈ 2.5, rounded to the nearest tenth.

Alternatively, you
could use natural
logarithms: .

Even without a calculator, we can approximate logarithms by finding the whole numbers between which a logarithm value lies. Logarithms of a given base are either always increasing or always decreasing as the argument of the logarithm (the x in logbx) increases. So, if the value of x is between q and r, then the value of logbx is between logbq and logbr.

To find which whole numbers log316 lies between, to check our answer to Example 13, find the powers of 3 that 16 lies between. The number 16 is between 32, or 9, and 33, or 27. Logarithms of base 3 increase for arguments from 9 to 27, so log316 is between log39 and log327, or log332 and log333. Simplifying these logarithms, we can see that log316 is between 2 and 3. Our answer of 2.5 makes sense.

The various logarithmic identities also come in handy for rewriting equations to solve for a single variable.

This is similar to the
process you would use
to estimate a square
root. For example, to
find the approximate
value of , you
would find the closest
two perfect squares
to 19. The number 19
is between 16 and 25,
so is between
and , or between 4 and 5.

Here is how you may see logarithmic identities on the ACT.

When x > 1, 5logxx−3 = ?

A. −15

B. −3/5

C. 3/5

D. 5/3

E. 2

Solve the following equation for t.

ln42t − 1 = 8

Start by using the power property, because this will pull the expression containing t out as a factor, which will make it easier to then isolate t.

(2t − 1) ⋅ ln4 = 8

Use the logarithmic power property.

2t ⋅ ln4 − ln4 = 8

Distribute the multiplication by ln4 to both terms in the binomial.

2t ⋅ ln4 = 8 + ln4

Add ln4 to both sides.

2t =

Divide both sides by ln4.

t =

Divide both sides by 2.

The answer may also be written as t = + 1/2. If we use a calculator, we can solve for the approximate value of t, 3.39.

Solve the following equation for x.

18 ⋅ 104x = 6

Start by dividing both sides by 18, to isolate the exponential expression containing x.

104x = 6/18

104x = 1/3

Simplify the fraction.

log104x = log(1/3)

Take the common logarithm of both sides.

4x = log(1/3)

Because the base is 10, you can use the identity logbbn = n.

x =

Divide both sides by 4.

Using your calculator, you can evaluate this expression to find that x ≈ −0.12.

The expression
log(1/3) is the same
as log3−1, which can
be rewritten using the
power property as
−1log3. Our answer
to Example 15 can be
written as x = ,
which again is equal
to about −0.12.

Solve the equation below, for k to the nearest tenth.

log4(8/k) = 5/4

Start by using the logarithmic quotient property to rewrite the left side of the equation.

log48 − log4k = 5/4

3/2 − log4k = 5/4

Evaluate log48.

1/4 − log4k = 0

Subtract 5/4 from both sides.

1/4 = log4k

Add log4k to both sides.

41/4 = k

Use the definition of logarithm.

k ≈ 1.4

Evaluate 41/4 on your calculator.

Perhaps you can
immediately recognize
that = 8, or
(41/2)3 = 8, so log48 =
3/2. But, if not, you
can rewrite log48 as
log423, and use the
logarithmic power
property to write it
as 3log42. The square
root (power of 1/2) of
4 is 2, so log42 = 1/2.
So, 3log42 = 3/2.

5logxx−3

= logx(x−3)5

Use the power property of logarithms.

= logxx−15

Raise x−3 to the power of 5: (x−3)5 = x−3 5.

= −15

Use the identity logbbn = n.

Alternatively, you could first evaluate logxx−3 and then multiply by 5.

5logxx−3

= 5 ⋅ (−3)

Use the identity logbbn = n.

= −15

Multiply.

The correct answer is (A).