Comparing Huge Powers Without a Calculator

Which is bigger, $2^{300}$ or $3^{200}$? Both numbers have around a hundred digits — actually computing either one by hand is out of the question, and even a calculator will choke on the exact digits. You don't need the exact values, though, just which one is bigger, and logarithms let you answer that from two ordinary-sized numbers instead.

The idea: logs turn exponent comparisons into multiplication

A logarithm is strictly increasing, meaning if $a > b$ then $\log a > \log b$, and the reverse holds too — so comparing $2^{300}$ and $3^{200}$ is exactly equivalent to comparing $\log(2^{300})$ and $\log(3^{200})$. Using the power rule $\log(x^n) = n\log x$:

$$\log(2^{300}) = 300 \log 2, \qquad \log(3^{200}) = 200 \log 3 .$$

In plain English: instead of comparing two astronomically large numbers directly, you compare two ordinary-sized numbers — an exponent times a small, memorizable log value — because the logarithm collapses exponentiation down to multiplication, and the comparison survives the transformation intact since log is increasing.

Worked example 1: 2^300 vs 3^200

Using $\log_{10} 2 \approx 0.301$ and $\log_{10} 3 \approx 0.477$ (worth memorizing): $300 \log 2 \approx 300 \times 0.301 = 90.3$, and $200 \log 3 \approx 200 \times 0.477 = 95.4$. Since $95.4 > 90.3$, $\log(3^{200}) > \log(2^{300})$, so $3^{200} > 2^{300}$ — decided from two quick multiplications instead of two hundred-digit numbers. As a sanity check, both logs also tell you roughly how many digits each number has: $10^{90.3}$ has 91 digits, $10^{95.4}$ has 96 digits, confirming $3^{200}$ isn't just bigger but noticeably so.

Worked example 2: a ratio-based shortcut without a log table

Compare $5^{44}$ and $4^{50}$ without any log values memorized, using a common trick: rewrite both as powers of the same exponent when possible. Here $44$ and $50$ share a factor of $2$: $5^{44} = (5^{22})^2$ and $4^{50} = (4^{25})^2$, so comparing $5^{44}$ vs $4^{50}$ reduces to comparing $5^{22}$ vs $4^{25}$ — smaller exponents, same conclusion since squaring preserves order for positive numbers. Repeating isn't always this clean, but when exponents share a large common factor, dividing it out first shrinks the problem before you even reach for logs. Failing that, logs always work: $\log(5^{44}) = 44\log5 \approx 44(0.699)=30.8$ and $\log(4^{50})=50\log4\approx50(0.602)=30.1$, so $5^{44} > 4^{50}$, a margin close enough that it's worth double-checking the arithmetic carefully.

The explorer above lets you drag the base and see how $\log_b(x)$ grows — the key fact this whole technique leans on is that it's strictly increasing, so it never reorders two numbers when you take logs of both, only rescales them to a size you can compare by hand.