Exponential and logarithmic functions

What are exponential and logarithmic functions?

In plain words: b^x grows faster than any power of x, and log_b undoes it, just as an inverse function undoes the original. NCERT’s comparison (page 126): at x = 10^³, x^¹⁰⁰ = 10^³⁰⁰, while 10^ˣ = 10^¹⁰⁰⁰.

Graphs and key facts

For a base b > 1 (NCERT pages 126–128):

• b^x has domain all real numbers and range the positive real numbers. Its graph passes through (0, 1), always rises, and approaches the x-axis on the left without meeting it.
• log_b x has domain the positive real numbers (there is no real logarithm of 0 or of a negative number) and range all real numbers. Its graph passes through (1, 0), always rises, and goes down along the y-axis as x → 0^⁺.

We cannot make a meaningful definition of logarithm of non-positive numbers and hence the domain of log function is R^⁺

NCERT Mathematics Part I (Class 12), Section 5.4, page 128

The graphs of e^x and log x are mirror images of each other in the line y = x (NCERT page 128), as the graphs of a function and its inverse always are.

Laws of logarithms

NCERT proves the product law and the change of base rule from the definition (page 128) and leaves the power and quotient laws to the reader (page 129). For a base b > 1 and positive p, q, x, y:

The change of base rule is log_a p = log_b p/log_b a.

For the product law: if b^β = p and b^γ = q, then pq = b^{β + γ}, so log_b pq = β + γ. For example, 625 = 5⁴ = 25², so log₅ 625 = 4 and log₂₅ 625 = 2 (NCERT page 127); by change of base, log₂₅ 625 = log₅ 625/log₅ 25 = 4/2.

NCERT slip (page 129): the two lines after the product law end “= 2 log p” and “= n log p”; with base b they should read 2 log_b p and n log_b p.

NCERT Example 25 (page 129): x = e^{log x} holds only for x > 0, because log x is defined only for positive x.

Derivatives of e^ˣ and log x

NCERT Theorem 5 (page 129; the chapter skips the proof and points to its supplementary material): e^x is its own derivative, and log x has derivative 1/x.

With the chain rule (NCERT Section 5.3.1) these handle composites. NCERT Example 26 (pages 129–130):

• d/dx (e^−x) = e^−x × (−1) = −e^−x;
• d/dx sin (log x) = cos (log x) × 1/x, for x > 0;
• d/dx cos^⁻¹ (e^x) = −e^x/√(1 − e^2x), which needs e^x < 1, that is x < 0;
• d/dx e^{cos x} = −sin x e^{cos x}.

Solved examples

Common mistakes to avoid

Quick revision summary

• b^x (b > 1) has domain R, range the positive reals and passes through (0, 1); log_b x has domain the positive reals, range R and passes through (1, 0). The graphs are mirror images in y = x.
• log x means log to base e; log (pq) = log p + log q, log (p/q) = log p − log q, log p^ⁿ = n log p, and log_a p = log_b p/log_b a.
• d/dx (e^ˣ) = e^ˣ and d/dx (log x) = 1/x; use the chain rule for composites.

Questions students ask

Why is the derivative of e^ˣ equal to e^ˣ?

The derivative of e^x is e^x by NCERT’s Theorem 5 (page 129), whose proof the chapter skips; the reason, from first principle, is that the difference quotient (e^{x + h} − e^x)/h equals e^x times (e^h − 1)/h, and e is the one base for which (e^h − 1)/h tends to 1 as h → 0. With h = 0.001 that factor is already about 1.0005, while for base 10 the same factor, (10^h − 1)/h, tends to log 10 ≈ 2.303, not 1.

Why is the derivative of log x equal to 1/x?

The derivative of log x is 1/x because log x is the inverse of e^x: writing y = log x as x = e^y and differentiating both sides with respect to x gives 1 = e^y dy/dx, so dy/dx = 1/e^y = 1/x, for x > 0. This is the method NCERT uses for sin^⁻¹ x, and like NCERT it assumes that the derivative exists. For negative x, log (−x) has the same derivative, 1/x, so d/dx log |x| = 1/x for every x ≠ 0, which is why ∫(1/x) dx = log |x| + C.

Is log x the same as ln x in Class 12 maths?

Yes, in NCERT’s Class 12 calculus log x is the natural logarithm, to base e, which calculators and many books write as ln x; NCERT says so for this chapter on page 127. A calculator’s log key, though, is base 10: log₁₀ 100 = 2, while ln 100 ≈ 4.605, and in general ln x = ln 10 × log₁₀ x ≈ 2.303 log₁₀ x. So read log x in a derivative as log_e x, with derivative 1/x, and handle a base-10 logarithm with the change of base rule, which brings in the factor 1/ln 10 ≈ 0.4343 (see Example 4).

What is the difference between exponential and logarithmic functions?

An exponential function b^x and the logarithmic function log_b x with the same base b > 1 are inverses: the exponential turns any real exponent into a positive number, and the logarithm turns that number back into the exponent, as in 4³ = 64 and log₄ 64 = 3. So their domains and ranges swap (all real numbers and the positive reals), and they grow at different rates: e^x outgrows every power of x, reaching about 22 026 at x = 10, while log x grows more slowly than every root, only about 2.303 at x = 10. Logarithms are the tool for solving exponential equations: 3^x = 20 gives x = log 20/log 3 ≈ 2.727.

Can the base of a logarithm be between 0 and 1?

Yes, a logarithm can have any positive base other than 1, so a base between 0 and 1 is allowed, though NCERT’s Definitions 3 and 4 (pages 126–127) take b > 1 only. With such a base both functions fall instead of rising: (1/2)^x decreases as x grows, and log_1/2 8 = −3 because (1/2)⁻³ = 8, so an inequality turns round when such a logarithm is removed: log_1/2 x > 2 means 0 < x < 1/4. The base 1 is excluded because 1^x = 1 for every x, which has no inverse, and negative bases fail because, for example, (−4)^1/2 is not a real number.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Differentiate sin (tan^⁻¹ e^−x) (NCERT Exercise 5.4, Q4, page 130).

   Show answerHide answer

   cos (tan^⁻¹ e^−x) × 1/(1 + e^−2x) × (−e^−x) = −e^−x cos (tan^⁻¹ e^−x)/(1 + e^−2x).

2. Q2. Differentiate e^x + e^x2 + … + e^x5 (NCERT Exercise 5.4, Q6, page 130).

   Show answerHide answer

   Term by term, the derivative of e^xk is kx^{k − 1}e^xk: e^x + 2xe^x2 + 3x²e^x3 + 4x³e^x4 + 5x⁴e^x5.

3. Q3. Differentiate √(e^√x), x > 0 (NCERT Exercise 5.4, Q7, page 130).

   Show answerHide answer

   √(e^√x) = e^√x/2, so the derivative is e^√x/2 × 1/(4√x). (NCERT writes the same as e^√x/(4√(xe^√x)).)

4. Q4. Differentiate cos (log x + e^x), x > 0 (NCERT Exercise 5.4, Q10, page 130).

   Show answerHide answer

   −sin (log x + e^x) × (1/x + e^x).