Logarithm rules — product, quotient, power, change of base with applications

Question

Simplify $\log_2 32 + \log_3 81 - \log_5 125$ . Then solve: if $\log_{10} x + \log_{10}(x-3) = 1$ , find $x$ .

(CBSE 11 & JEE Main pattern)

Solution — Step by Step

Product rule: $\log_a(mn) = \log_a m + \log_a n$
Quotient rule: $\log_a(m/n) = \log_a m - \log_a n$
Power rule: $\log_a(m^n) = n\log_a m$
Change of base: $\log_a b = \dfrac{\log_c b}{\log_c a}$

Special values: $\log_a a = 1$ , $\log_a 1 = 0$

$\log_2 32 = \log_2 2^5 = 5$

$\log_3 81 = \log_3 3^4 = 4$

$\log_5 125 = \log_5 5^3 = 3$

5 + 4 - 3 = \mathbf{6}

\log_{10} x + \log_{10}(x-3) = 1

Using the product rule:

\log_{10}[x(x-3)] = 1

Converting to exponential form:

x(x-3) = 10^1 = 10

x^2 - 3x - 10 = 0

(x-5)(x+2) = 0

$x = 5$ or $x = -2$

For $\log_{10} x$ to exist, $x > 0$ . For $\log_{10}(x-3)$ to exist, $x > 3$ .

So $x = -2$ is rejected. Answer: $x = 5$ .

Why This Works

Logarithms convert multiplication to addition (product rule), division to subtraction (quotient rule), and powers to multiplication (power rule). This is why they were invented — to simplify calculations before calculators existed.

graph TD
    A["Log Simplification"] --> B{"What operation?"}
    B -->|"log(m × n)"| C["Split: log m + log n"]
    B -->|"log(m / n)"| D["Split: log m - log n"]
    B -->|"log(m^n)"| E["Bring down: n × log m"]
    B -->|"Different bases"| F["Change base:<br/>log_a b = log b / log a"]
    A --> G["Log Equation"]
    G --> H["Combine using rules"]
    H --> I["Convert to exponential form"]
    I --> J["Solve algebraically"]
    J --> K["CHECK: domain restrictions<br/>argument must be > 0"]

Alternative Method — Convert Everything to Same Base

For the equation, we could also use natural logs:

\ln x + \ln(x-3) = \ln 10

\ln[x(x-3)] = \ln 10

x(x-3) = 10

Same quadratic, same answer. The base doesn’t matter for equations — it cancels out.

For JEE: the change of base formula has a powerful consequence: $\log_a b \times \log_b c = \log_a c$ . This “chain rule” for logs appears in many JEE problems. Also, $\log_a b = 1/\log_b a$ — useful for flipping bases quickly.

Common Mistake

Students forget to check domain restrictions after solving log equations. The argument of a logarithm must be strictly positive. In our problem, $x = -2$ gives $\log_{10}(-2)$ , which is undefined. Always verify that your solutions make the original log expressions valid. Losing marks for including extraneous roots is one of the most common errors in JEE Main.

Question

Solution — Step by Step

Why This Works

Alternative Method — Convert Everything to Same Base

Common Mistake

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