Logarithms — From Confusion to Clarity

Logarithms exist because exponentiation creates big numbers very fast. We needed a way to invert this process. If $2^{10} = 1024$ , then $\log_2 1024 = 10$ — the logarithm answers the question “what power did I raise the base to?”

Before calculators, logarithms were the standard tool for multiplying large numbers. Today, they’re essential for understanding exponential growth, information theory, pH and decibels, and for solving equations where the unknown is in the exponent.

Most students find logs hard because they try to memorise rules before understanding what a log actually is. We’ll fix that.

Key Terms and Definitions

Logarithm: $\log_a b = c$ means $a^c = b$ . Read as “log base $a$ of $b$ equals $c$ .”

Base: The number $a$ being raised to a power. Must be positive and not equal to 1.

Argument: The number $b$ you’re taking the log of. Must be positive.

Common logarithm: $\log_{10}$ , written as just $\log$ . Used in pH, decibels, Richter scale.

Natural logarithm: $\log_e = \ln$ . Base $e \approx 2.718$ . Used in calculus, compound interest.

Antilogarithm: The inverse of the logarithm. $\text{antilog}(x) = 10^x$ .

The Core Identity (Never Forget This)

\log_a b = c \iff a^c = b

Every log property follows from this one identity. If you ever forget a rule, you can re-derive it.

All the Rules, Derived

\log_a(mn) = \log_a m + \log_a n \quad \text{(Product Rule)}

\log_a\left(\frac{m}{n}\right) = \log_a m - \log_a n \quad \text{(Quotient Rule)}

\log_a(m^n) = n \log_a m \quad \text{(Power Rule)}

\log_a a = 1 \quad \text{(Base Rule)}

\log_a 1 = 0 \quad \text{(Zero Rule)}

\log_a b = \frac{\log_c b}{\log_c a} \quad \text{(Change of Base Rule)}

Why the product rule works: Let $\log_a m = p$ and $\log_a n = q$ . Then $m = a^p$ and $n = a^q$ . So $mn = a^p \cdot a^q = a^{p+q}$ . Therefore $\log_a(mn) = p + q = \log_a m + \log_a n$ . The other rules follow similarly.

Methods for Solving Log Equations

Method 1: Convert to Exponential Form

$\log_3 x = 4 \implies x = 3^4 = 81$ .

Any time you have " $\log_a(\text{expression}) = \text{number}$ ", convert directly.

Method 2: Use Properties to Simplify

$\log x + \log(x-3) = 1 \implies \log[x(x-3)] = 1 \implies x(x-3) = 10 \implies x^2 - 3x - 10 = 0$

Solve the quadratic: $(x-5)(x+2) = 0 \implies x = 5$ or $x = -2$ .

Check: $x = -2$ would give $\log(-2)$ which is undefined. So $x = 5$ only.

Method 3: Change of Base for Mixed Bases

$\log_2 x = \log_4(x + 6)$

Change both to base 2: $\log_4(x+6) = \frac{\log_2(x+6)}{\log_2 4} = \frac{\log_2(x+6)}{2}$ .

So: $\log_2 x = \frac{\log_2(x+6)}{2} \implies 2\log_2 x = \log_2(x+6) \implies \log_2 x^2 = \log_2(x+6)$

$\implies x^2 = x + 6 \implies x^2 - x - 6 = 0 \implies (x-3)(x+2) = 0 \implies x = 3$ (rejecting $x = -2$ ).

Solved Examples

Easy (CBSE Class 10): Evaluate using rules

Q: Find the value of $\log_5 500 - \log_5 4$ .

$\log_5 500 - \log_5 4 = \log_5(500/4) = \log_5 125 = \log_5 5^3 = 3$ . ✓

Medium (JEE Main level): Simplify

Q: Simplify $\log_2 \sqrt{8} + \log_4 2$ .

$\log_2 \sqrt{8} = \log_2 8^{1/2} = \frac{1}{2}\log_2 8 = \frac{1}{2} \times 3 = \frac{3}{2}$ .

$\log_4 2 = \frac{\log_2 2}{\log_2 4} = \frac{1}{2}$ .

Total: $\frac{3}{2} + \frac{1}{2} = 2$ .

Hard (JEE Advanced level): Proving an inequality

Q: If $a > 1$ , prove that $\log_a\left(\frac{a+1}{a-1}\right) < \frac{2}{a^2-1}$ .

This requires the inequality $\ln(1+x) < x$ for $x > 0$ . Let $x = \frac{2}{a-1}$ . With natural log:

$\ln\left(\frac{a+1}{a-1}\right) = \ln\left(1 + \frac{2}{a-1}\right) < \frac{2}{a-1}$ (using $\ln(1+x) < x$ ).

Dividing by $\ln a > 0$ : $\log_a\left(\frac{a+1}{a-1}\right) < \frac{2}{(a-1)\ln a}$ .

And since $\ln a < a - 1$ for $a > 1$ (another standard inequality), $\frac{1}{\ln a} > \frac{1}{a-1}$ , giving $\frac{2}{(a-1)\ln a} > \frac{2}{(a-1)^2} > \frac{2}{a^2-1}$ for large $a$ . (A complete proof requires more care, but the log inequality technique is what JEE tests.)

Exam-Specific Tips

JEE Main: Log questions appear in the “sequences and series” section and as standalone algebra. The change of base formula and manipulation of expressions like $\log_2 3 \times \log_3 4$ (answer = $\log_2 4 = 2$ ) are common. Also: $\log_a b \times \log_b a = 1$ .

CBSE Class 11: Chapter 2 (Relations and Functions) introduces logarithmic functions. Know the graphs of $y = \log_a x$ for $a > 1$ (increasing) and $0 < a < 1$ (decreasing).

NEET: Logarithms appear indirectly in pH calculations: $\text{pH} = -\log[\text{H}^+]$ . If $[\text{H}^+] = 10^{-7}$ , then pH = 7. Know how to convert between exponential and log form quickly.

Common Mistakes to Avoid

Mistake 1: " $\log(a + b) = \log a + \log b$ " — WRONG. The product rule says $\log(ab) = \log a + \log b$ . There’s no rule for $\log(a + b)$ .

Mistake 2: " $\log_a b^2 = (\log_a b)^2$ " — WRONG. The power rule gives $\log_a b^2 = 2\log_a b$ . The square is on the argument, not the result.

Mistake 3: Not checking for domain validity. After solving a log equation, always check that the argument of every log in the original equation is positive. Reject any solution that makes an argument ≤ 0.

Mistake 4: Confusing $\log_a$ with $\log_{10}$ . In most Indian textbooks, $\log$ without a base means $\log_{10}$ (common log). In calculus and engineering, $\ln$ = natural log. Don’t mix them up.

Mistake 5: " $\frac{\log m}{\log n} = \log\frac{m}{n}$ " — WRONG. Change of base says $\frac{\log m}{\log n} = \log_n m$ — the ratio of logs is a log with a different base, not a log of the ratio.

Practice Questions

Q1: Evaluate $\log_{10} 1000$ .

$\log_{10} 1000 = \log_{10} 10^3 = 3$ .

Q2: Simplify $\log_2 32 - \log_2 4$ .

$\log_2(32/4) = \log_2 8 = 3$ .

Q3: Find $x$ if $\log_3 x = 4$ .

$x = 3^4 = 81$ .

Q4: Simplify $\log_a a^5$ .

$5 \log_a a = 5 \times 1 = 5$ .

Q5: If $\log 2 = 0.301$ and $\log 3 = 0.477$ , find $\log 72$ .

$\log 72 = \log(8 \times 9) = \log 2^3 + \log 3^2 = 3(0.301) + 2(0.477) = 0.903 + 0.954 = 1.857$ .

Q6: Solve $\log(x+2) + \log(x-1) = 1$ .

$\log[(x+2)(x-1)] = 1 \implies (x+2)(x-1) = 10 \implies x^2 + x - 12 = 0 \implies (x+4)(x-3) = 0$ . $x = 3$ (reject $x = -4$ as it makes $\log(-2)$ undefined).

Q7: Find the value of $\log_4 8$ .

$\log_4 8 = \frac{\log 8}{\log 4} = \frac{3\log 2}{2\log 2} = \frac{3}{2}$ .

Q8: Is $\log_2 3$ rational or irrational? Why?

Irrational. If $\log_2 3 = p/q$ (rational), then $3 = 2^{p/q}$ , so $3^q = 2^p$ . But this contradicts the fundamental theorem of arithmetic (a power of 2 cannot equal a power of 3). So $\log_2 3$ is irrational.

FAQs

Q: Why is $\log 0$ undefined? Because $a^c = 0$ has no solution for any finite $c$ and $a > 0$ . The function $a^c$ is always positive — it can approach 0 but never reach it.

Q: Why is $\log$ of a negative number undefined (in real numbers)? Because $a^c > 0$ for all real $c$ and $a > 0$ . There’s no real power you can raise a positive number to and get a negative result.

Q: What’s the difference between $\log$ and $\ln$ ? $\log = \log_{10}$ (common log), $\ln = \log_e$ (natural log, $e \approx 2.718$ ). They’re related by $\ln x = \log x / \log e \approx 2.303 \log x$ .

Q: How do I use log tables? Express the number in scientific notation: $N = m \times 10^n$ where $1 \leq m < 10$ . Then $\log N = n + \log m$ . Look up $\log m$ in the four-figure log table. This is still tested in some CBSE board exams.

Advanced Applications

Logarithmic equations with conditions

When solving $\log_a f(x) = g(x)$ , always check:

Base $a > 0$ , $a \neq 1$
Argument $f(x) > 0$
The solution satisfies the original equation

These domain checks reject extraneous solutions that algebraic manipulation introduces.

Logarithmic inequalities

The direction of inequality flips depending on the base:

If $a > 1$ : $\log_a x > \log_a y \iff x > y$ (both positive)
If $0 < a < 1$ : $\log_a x > \log_a y \iff x < y$ (inequality flips!)

Forgetting to flip the inequality when the base is between 0 and 1 is one of the most common JEE errors. Always check the base before manipulating log inequalities.

Logarithmic differentiation preview

In calculus, $\frac{d}{dx}(\ln x) = \frac{1}{x}$ and $\frac{d}{dx}(\log_a x) = \frac{1}{x \ln a}$ .

Logarithmic differentiation is used for functions like $y = x^x$ : take $\ln$ of both sides, then differentiate. This technique relies entirely on fluency with log properties.

Additional Solved Examples

Hard (JEE Main): Find the number of digits in $2^{100}$ . Given $\log 2 = 0.301$ .

$\log(2^{100}) = 100 \times 0.301 = 30.1$

The characteristic is 30, so $2^{100}$ has $30 + 1 = \mathbf{31}$ digits.

Medium: Solve $\log_3(x+2) - \log_3(x-2) = 2$ .

$\log_3\frac{x+2}{x-2} = 2 \implies \frac{x+2}{x-2} = 9 \implies x + 2 = 9x - 18 \implies 8x = 20 \implies x = 2.5$ .

Check: $x - 2 = 0.5 > 0$ . Valid.

Additional Practice Questions

Q9. If $\log_2(x-1) + \log_2(x+1) = 3$ , find $x$ .

$\log_2[(x-1)(x+1)] = 3 \implies x^2 - 1 = 8 \implies x^2 = 9 \implies x = 3$ (reject $x = -3$ since $\log_2(-4)$ is undefined).

Q10. How many digits does $5^{20}$ have? (Use $\log 5 = 0.699$ )

$\log(5^{20}) = 20 \times 0.699 = 13.98$ . Characteristic = 13. Number of digits = $13 + 1 = 14$ .

Logarithmic and Exponential Functions — Graphs

The graph of $y = \log_a x$ (for $a > 1$ ) is:

Defined only for $x > 0$
Passes through $(1, 0)$ always ( $\log_a 1 = 0$ )
Passes through $(a, 1)$ ( $\log_a a = 1$ )
Increasing (slope positive but decreasing)
Concave down
Approaches $-\infty$ as $x \to 0^+$

For $0 < a < 1$ , the graph is decreasing (reflected about the x-axis compared to $a > 1$ ).

The exponential function $y = a^x$ is the inverse — its graph is the reflection of $\log_a x$ across $y = x$ .

Logarithms in chemistry — pH

\text{pH} = -\log_{10}[\text{H}^+]

If $[\text{H}^+] = 10^{-3}$ M, then pH $= 3$ (acidic). If $[\text{H}^+] = 10^{-11}$ M, then pH $= 11$ (basic). Neutral water: $[\text{H}^+] = 10^{-7}$ , pH $= 7$ .

Logarithms in physics — decibels

\text{dB} = 10\log_{10}\frac{I}{I_0}

where $I_0 = 10^{-12}$ W/m $^2$ (threshold of hearing). A sound of intensity $10^{-6}$ W/m $^2$ has level $10\log(10^6) = 60$ dB (normal conversation).

These cross-subject applications of logarithms appear in both Chemistry (pH) and Physics (decibels) papers. The underlying skill is identical: converting between exponential and logarithmic form.