Laws of Logs

Law of logs proof

It is not necessary to be able to prove each logarithm law for the exam, but it is important to understand each step and why it occurs.

Product (addition) law

1. If \(\log_x(a) = c\) and \(\log_x(b) = d\), then you can rewrite the logarithms as an exponential function.

x, the exponent is c, the answer to the exponential is a.

Therefore, it can be written as \(x^c = a\)

For \(\log_x(b) = d\), the base is x, the exponent is d, and the answer to the exponential is b.

Therefore, it can be written as \(x^d = b\)

2. Thus, using our exponential (indices) rule of \(M^n \cdot M^n = M^{m+n}\),

\(ab = (x^c)(x^d) = x^{c+d}\)

\(ab = x^{c+d}\)

3. Take the log of both sides:

\(\log_x(ab) = \log_x(x^{c+d})\)

4. Because \(\log_x(x^{c+d})\) includes both an exponential with a base of x and a logarithm with a base of x (\(\log_x(x^{c+d})\)), they will cancel each other out to become just c + d.

\(log_x(x^{c+d}) = c +d \)

\(\log_x(ab) = c+d\)

5. We defined c and d in part 1. \(\log_x(a) = c\) and \(\log_x(b) = d\)

Therefore, \(c +d = \log_x(a) + \log_x(b)\)

\(\log_x(ab) = \log_x(a) + \log_x(b)\)

Quotient rule

1. If \(\log_x(a) = c\) and \(\log_x(b) = d\), then you can rewrite the logarithms as an exponential function.

For\(\log_x(a) = c\), the base is x, the exponent is c, and the answer to the exponential is a.

Therefore, it can be written as \(x^c = a\)

For \(\log_x(b) = d\), the base is x, the exponent is d, and the answer to the exponential is b.

Therefore, it can be written as \(x^d = b\)

2. Thus, using our exponential (indices) rules of \(\frac{M^m}{M^n} = M^{m-n}\),

\(\frac{a}{b} = \frac{x^c}{x^d} = x^{c-d}\)

\(\frac{a}{b} = x^{c-d}\)

3. Take the log of both sides.

\(\log_x(\frac{a}{b}) = \log_x(x^{c-d})\)

4. Because \(\log_x(x^{c-d})\) includes both an exponential with a base of x and a logarithm with a base of x, they will cancel each other out to become just c - d .

\(\log_x(x^{c-d}) = c-d\)

\(\log_x(\frac{a}{b}) = c-d\)

5. We defined c and d in part 1 where \(\log_x(a) = c\) and \(\log_x(b) = d\):

\(c-d = \log_x(a) - \log_x(b)\)

\(\log_x(\frac{a}{b}) = \log_x(a) - \log_x(b)\)

Change of base

1. Let \(\log_a(x) = k\) where the base is a ,the exponent is k, and the answer to the exponential = x.

Therefore, it can be rewritten as an exponential: \(a^k = x\)

2. Take the log of both sides

\(\log_b(a^k) = \log_b(x)\)

3. Use the power rule to simplify

\(\log_b(a^k) = k\log_b(a)\) which you can then substitute back into the equation

\(k\log_b(a) = \log_b(x)\)

4. Rearrange to get k on its own by dividing through k \(\log_b(a)\)

\(k = \frac{\log_b(x)}{\log_b(a)}\)

5. As k is already defined, it can be substituted into the equationk\(\log_a(x)\)

\(\log_a(x) = \frac{\log_b(x)}{\log_b(a)}\)

Reciprocal law

1. \(\log_a(\frac{1}{x})\) can be written as \(\log_a(x^{-1})\) using our exponential rules with negatives.
2. You can use the power log rule to bring the -1 down so \(\log_a(x^{-1})\) becomes \(-\log_a(x)\).

Log of the base

1. Set \(\log_a(a)=x\) where the base is a, the exponent is x, and the answer to the exponential is a. Therefore, it can be written as\(a^x = a\).
2. According to exponential rules, if the answer of an exponential is equal to the base, then the exponent must be 1.

Log of 1

1. Set \(\log_a1=x\) where the base is a, the exponent is x, and the answer to the exponential is 1. Therefore, it can be written as\(a^x = 1\).
2. According to exponential rules, if the answer of an exponential is 1, then the exponent must be 0.

Simplifying and solving using laws of logs

Here, we will go through some examples of simplifying a range of laws of logs.

Simplifying and solving using 1 log law

Simplifying and solving using multiple log laws

It might help to use rules that simplify individual logs before doing the simplifying multiple log laws.