Expanding Logarithms | ChiliMath

Expanding Logarithms

When you are asked to expand log expressions, your goal is to express a single logarithmic expression into many individual parts or components. This process is the exact opposite of condensing logarithms because you compress a bunch of log expressions into a simpler one.

The best way to illustrate this concept is to show a lot of examples. In this lesson, there are eight worked problems.

The key to successfully expanding logarithms is to carefully apply the rules of logarithms. Take time to go over the rules and understand what they are trying to “say”.

For instance, Rule 1 is called the Product Rule. What it does is break the product of expressions as a sum of log expressions. See the rest of the descriptions below.

Rules of Logarithms

The seven rules of logarithms.

Study the description of each rule to get an intuitive understanding of it.

Descriptions of Logarithm Rules

Rule1: Product Rule

log base b of (MN) = log base b of (M) + log base b of (N)

The logarithm of the product of numbers is the sum of the logarithms of individual numbers.

Rule 2: Quotient Rule

log base b of (M/N) = log base b of (M) - log base b of (N)

The logarithm of the quotient of numbers is the difference of the logarithm of individual numbers.

Rule 3: Power Rule

log base b of (M^k) = (k) times

The logarithm of an exponential number is the exponent times the logarithm of the base.

Rule 4: Zero Rule

log base b of 1 = 0 where b>0

The logarithm of 1 with b > 0 but b \ne 1 equals zero.

Rule 5: Identity Rule

log base b of b = 1, where b > 1

The logarithm of a number that is equal to its base is just 1. Since the number is also the base, b, that means b>0 but b \ne 1.

Rule 6: Log of Exponent Rule

log base b of (b^k) = k

The logarithm of an exponential number where its base is the same as the base of the log equals the exponent.

Rule 7: Exponent of Log Rule

b^(log base b of k) = k

Raising the logarithm of a number by its base equals the number.

Examples of How to Expand Logarithms

Example 1: Expand the log expression.

log base 5 of 3xyk

Looking inside the parenthesis, we see a product of a number and variables. The Product Rule doesn’t say that there should only be two factors inside, in fact, there could be more. Okay, so we’ll separate the main log expression as the sum of four logs.

log base 5 of 3xyk is equal to log base 5 of (3) plus log base 5 of (x) plus log base 5 of (y) plus log base 5 of (k)

Example 2: Expand the log expression.

log base 2 of 7x over y

The inside of the parenthesis is a fraction which means I will first apply the Quotient Rule. Since the numerator is a product of 7 and x, I use the Product Rule to break it up.

log base 2 of 7x over y is equal to log base 2 of (7) plus log base 2 of (x) minus log base 2 of (y)

Example 3: Expand the log expression.

log base 6 of 36 m cubed over the sqrt of n

Okay, this one is also in fraction form so the Quotient Rule is the first step that we’re going to apply.

Now, there’s something “new” in this problem. That is, the numerator contains a variable with an exponent. This should be easy since Rule 3 or Power Rule can easily handle it. Just bring the exponent down to the left, that’s it!

In addition, there’s a radical expression in the denominator. Remember that a radical can be expressed as a fractional exponent. Since this radical is a square root that means the power is just \large{1 \over 2}.

log base 6 of 36 m cubed over the sqrt of n is equal to 2 plus 3 times log base 6 of m minus one half of log base 6 of n

Also, this is the first time we see Rule 5 or Identity Rule of Logarithm in action. Expect to apply this rule more often because it is extremely useful in the simplification process.

Example 4: Expand the log expression.

log base 2 of sqrt 7 x cubed y

Don’t let that square root symbol scare you. Just think of it as the power or exponent of \large{1 \over 2}. So this problem is reduced to expanding a log expression with a power of \large{1 \over 2}. This is where the Power Rule brings down that exponent \large{1 \over 2} to the left of the log, and then you expand the rest as usual.

log base 2 of sqrt 7 x cubed y is equal to one half of the quantity log base 2 of 7 plus 3 times log base 2 of x plus log base 2 of y

Example 5: Expand the log expression.

log base 7 of the quantity x sqrt of y over h, raised to the 4th power

This problem is quite interesting because the entire expression is being raised to some power. In addition, the presence of a square root on the numerator adds some level of difficulty.

However, if we stick to the basics by carefully applying the rules of exponents in every step, we should have no problem conquering this problem! Let’s plan out our course of action, okay?

Bring down that exponent 4 using the Power Rule. Next, use the Quotient Rule to express the fraction as a difference of log expressions. And finally, don’t forget that square root is just a fractional exponent of \large{1 \over 2}.

log base 7 of the quantity x sqrt of y over h, raised to the 4th power is equal to 4 times log base 7 of x plus 2 times log base 7 of y minus 4 times log base 7 of h

I distributed the 4 into the grouping symbol to get rid of the fraction \large{1 \over 2}. It’s also okay to leave the 4 outside the grouping symbol which means we don’t have to distribute the 4 into the expressions inside the square brackets. Either of the two answers should be correct.

Example 6: Expand the log expression.

log base 10 of the cube root of m over 5yk

It’s just a cube root of some rational expression. Replace the cube root symbol by a fractional power of \large{1 \over 3}. That exponent of \large{1 \over 3} can be brought down using the Power Rule of Logarithm.  Now we have just to deal with the rational expression using the Quotient Rule, then finish it off using the Product Rule.

log base 10 of the cube root of m over 5yk is equal to one-third of the quantity log base 10 of m minus log base 10 of 5 minus log base 10 of y minus log base 10 of k

Don’t get distracted by the different types of grouping symbols. The idea is to make sure that we are applying the logarithm rules correctly in each step that we undertake, without committing algebraic mistakes such as distributing -1 into the grouping symbol.

Example 7: Expand the log expression.

log base 4 of 64 times the sqrt of x times the sqrt of y

I would immediately apply the Product Rule to separate the factors into a sum of logarithmic terms. Numerically simplify the {\log _4}64 since 64 can be expressed as an exponential number with base 4, that is 64 = {4^3}. Deal with the square roots by replacing them with fractional power then use the Power Rule of logarithms to bring it down in front of the log symbol as a multiplier.

log base 4 of 64 times the sqrt of x times the sqrt of y is equal to 3 plus one half of log base 4 of x plus one half of log base 4 of y

Example 8: Expand the log expression.

log base 2 of the quantity 4 times sqrt of y over the cube root of x raised to the third power.

The main power 3 can be placed out front as a multiplier using the Power Rule. The Quotient rule should deal with the fractional expressions by writing them as the difference of logs. Now, replace the square root symbol with the exponent of \large{1 \over 2}. Similarly, replace the cube root symbol with the exponent of \large{1 \over 3}.

log base 2 of the quantity 4 times sqrt of y over the cube root of x raised to the third power is equal to 6 plus 3 over 2 times the log base 2 of y minus log base 2 of x

You might also be interested in:

Condensing Logarithms

Logarithm Explained

Logarithm Rules

Solving Logarithmic Equations

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