Logarithmic Function and Its Graph: PECTAA Class 12 Maths, Punjab Board

Take the exponential function y = a^x and swap x and y. What you get is the logarithmic function. Every fact about its graph comes from that one swap.

Section 1.6 of Unit 1 in the PECTAA Mathematics 12 book is the Logarithmic Function. This note covers it for FSc Part 2 (2nd year, Class 12) students in Lahore, Faisalabad, Sargodha, Rawalpindi, Multan, Gujranwala and the other Punjab boards. Gujrat district comes under the Gujranwala board. PECTAA writes the books. Your own board sets your paper.

What to know

Definition. A logarithmic function is defined by y = log_a x, where a > 0, a is not equal to 1, and x > 0. It gives each positive real number x a unique real number y such that a^y = x.

Inverse of the exponential function. The exponential function y = a^x (a > 0, a not equal to 1) is one-to-one. Its domain is ]-infinity, infinity[ and its range is ]0, infinity[. So it can be inverted. Swapping x and y gives x = a^y, which means y = log_a x. Hence the logarithmic function has domain ]0, infinity[ and range ]-infinity, infinity[.

Special bases. A logarithm with base 10 is the common logarithm, written log b. A logarithm with base e is the natural logarithm, written ln b.

Basic identities (a > 0, a not equal to 1, x > 0):

• log_a 1 = 0
• log_a a = 1
• log_a (a^x) = x
• a^(log_a x) = x

Laws and change of base (x > 0, y > 0, t real):

1. Product law: log_a (xy) = log_a x + log_a y
2. Quotient law: log_a (x/y) = log_a x - log_a y
3. Power law: log_a (x^t) = t log_a x
4. Change of base: log_a x = (log_b x) / (log_b a). In particular log_a x = (ln x) / (ln a) = (log x) / (log a).

Graph behaviour (sections 1.6.1 and 1.6.2).

1. One-to-one. If log_a s = log_a t, then a^(log_a s) = a^(log_a t), so s = t.
2. Vertical asymptote x = 0. For a > 1, as x tends to 0 from the right, log_a x tends to -infinity. For 0 < a < 1, it tends to +infinity. Since the function is defined only for x > 0, there is no other asymptote.
3. Increasing or decreasing. Take x1 < x2. Put y1 = log_a x1 and y2 = log_a x2, so x1 = a^(y1) and x2 = a^(y2). For a > 1 this gives y1 < y2, so the function is strictly increasing. For 0 < a < 1 it gives y1 > y2, so it is strictly decreasing.
4. Intercepts. Put y = 0: log_a x = 0 gives x = a^0 = 1. The graph meets the x-axis at (1, 0). There is no y-intercept.
5. Continuity. The function is continuous for all x > 0, with no breaks or jumps.

Key points

Figure 1.10 in the book shows graphs with bases greater than 1, including ln x and log x. Figure 1.11 shows bases between 0 and 1. All the curves pass through (1, 0).

Link to the exponential graph. The book makes the logarithm the inverse of a^x. Their domain and range swap places, which is why the log graph lies to the right of the y-axis.

Worked examples

These use the book laws. The arithmetic is checked.

Example 1. Find log_2 8.

1. Write 8 = 2^3.
2. log_2 (2^3) = 3 by the identity log_a (a^x) = x.
3. So log_2 8 = 3.

Example 2. Find log 0.01.

1. The base is 10, so write 0.01 = 10^(-2).
2. log (10^(-2)) = -2.

Example 3. Simplify log 100 + log 1000.

1. Product law: log 100 + log 1000 = log (100 x 1000) = log 100000.
2. 100000 = 10^5, so the answer is 5.
3. Check: 2 + 3 = 5.

Example 4. Find log_4 32 by change of base.

1. log_4 32 = (ln 32) / (ln 4).
2. 32 = 2^5 and 4 = 2^2, so this is 5 ln 2 / 2 ln 2 = 5/2.
3. The answer is 2.5.

Example 5. Find log_0.5 8.

1. Use change of base: log_0.5 8 = (ln 8) / (ln 0.5).
2. ln 8 = 3 ln 2 and ln 0.5 = -ln 2.
3. So the value is 3 ln 2 / (-ln 2) = -3.
4. The base is between 0 and 1, so the function is decreasing. Since 8 > 1, the value is negative.

Common mistakes in the exam

1. Using a = 1 or a negative base. The definition needs a > 0 and a not equal to 1.
2. Taking log of zero or a negative number. The domain is x > 0 only.
3. Writing log (x + y) = log x + log y. The product law is for log (xy).
4. Saying the graph cuts the y-axis. The function is undefined at x = 0.
5. Mixing up increasing and decreasing. Check whether the base is above 1 or between 0 and 1.
6. Forgetting the change-of-base ratio. It is log_b x over log_b a, not the other way round.

Exam technique

• Start with the definition and say which condition on a you use.
• For a graph question, mark the point (1, 0) and the line x = 0 first.
• State the base before saying "increasing" or "decreasing".
• Check your board's notice for the sections in your paper.

Practice questions

1. State the domain and range of y = log_a x. Answer: Domain ]0, infinity[ and range ]-infinity, infinity[.
2. Find log_3 81. Answer: 81 = 3^4, so the answer is 4.
3. Find log_2 64 - log_2 2. Answer: By the quotient law this is log_2 32 = 5. Check: 6 - 1 = 5.
4. Where does the graph of y = log_a x cut the x-axis? Answer: At the point (1, 0).
5. Is y = log_0.5 x increasing or decreasing? Answer: Decreasing, because the book proves that log_a x is strictly decreasing when 0 < a < 1.

Ready to practise this? Continue in 2nd Year Mathematics.