Evaluate log₂(32) without a calculator

Question

Evaluate $\log_2 32$ without a calculator.

Solution — Step by Step

Let $\log_2 32 = x$ . By the definition of logarithm, this means:

2^x = 32

32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^5

So the equation becomes:

2^x = 2^5

Since the bases are equal, the exponents must be equal:

x = 5

Therefore: $\log_2 32 = \mathbf{5}$

Why This Works

The logarithm $\log_b a = x$ asks the question: “To what power must we raise $b$ to get $a$ ?” Rewriting in exponential form ( $b^x = a$ ) often makes the answer obvious, especially when $a$ is a perfect power of $b$ .

The key skill is recognising common powers: $2^1=2$ , $2^2=4$ , $2^3=8$ , $2^4=16$ , $2^5=32$ , $2^6=64$ , $2^7=128$ , $2^8=256$ , $2^9=512$ , $2^{10}=1024$ .

Alternative Method — Using Logarithm Properties

\log_2 32 = \log_2(2^5) = 5 \cdot \log_2 2 = 5 \times 1 = 5

Using the power rule: $\log_b(b^n) = n$ , so $\log_2(2^5) = 5$ .

Common Mistake

Students sometimes try to compute $32/2 = 16$ and think " $\log_2 32 = 16$ " or similar — confusing division with logarithms. Logarithm asks “what is the exponent?” not “how many times does 2 go into 32 by division?” Always convert to exponential form first: $\log_2 32 = x \Leftrightarrow 2^x = 32$ .

Build a quick lookup table for powers of 2 in your head. Powers of 2 appear constantly in logarithm, binary (computer science), and combinatorics problems. The pattern: $2, 4, 8, 16, 32, 64, 128, 256, 512, 1024$ corresponds to exponents $1$ through $10$ . Memorise these.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Logarithm Properties

Common Mistake

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