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Foundational

Comparing Huge Powers Without a Calculator

To decide which of two enormous exponentials is larger without computing either, take logarithms of both, logs turn an impossible multiplication into an easy comparison of two ordinary numbers.

Which is bigger, 23002^{300} or 32003^{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>ba > b then loga>logb\log a > \log b, and the reverse holds too, so comparing 23002^{300} and 32003^{200} is exactly equivalent to comparing log(2300)\log(2^{300}) and log(3200)\log(3^{200}). Using the power rule log(xn)=nlogx\log(x^n) = n\log x:

log(2300)=300log2,log(3200)=200log3.\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 log1020.301\log_{10} 2 \approx 0.301 and log1030.477\log_{10} 3 \approx 0.477 (worth memorizing): 300log2300×0.301=90.3300 \log 2 \approx 300 \times 0.301 = 90.3, and 200log3200×0.477=95.4200 \log 3 \approx 200 \times 0.477 = 95.4. Since 95.4>90.395.4 > 90.3, log(3200)>log(2300)\log(3^{200}) > \log(2^{300}), so 3200>23003^{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: 1090.310^{90.3} has 91 digits, 1095.410^{95.4} has 96 digits, confirming 32003^{200} isn't just bigger but noticeably so.

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

Compare 5445^{44} and 4504^{50} without any log values memorized, using a common trick: rewrite both as powers of the same exponent when possible. Here 4444 and 5050 share a factor of 22: 544=(522)25^{44} = (5^{22})^2 and 450=(425)24^{50} = (4^{25})^2, so comparing 5445^{44} vs 4504^{50} reduces to comparing 5225^{22} vs 4254^{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(544)=44log544(0.699)=30.8\log(5^{44}) = 44\log5 \approx 44(0.699)=30.8 and log(450)=50log450(0.602)=30.1\log(4^{50})=50\log4\approx50(0.602)=30.1, so 544>4505^{44} > 4^{50}, a margin close enough that it's worth double-checking the arithmetic carefully.

Function explorer
-2222.8
x = 1.00f(x) = 0.000

The explorer above lets you drag the base and see how logb(x)\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.

To compare two large exponentials ama^m vs bnb^n, take logs: compare mlogam \log a vs nlogbn\log b, two ordinary multiplications instead of two unmanageable numbers. Memorizing log1020.301\log_{10}2\approx0.301, log1030.477\log_{10}3\approx0.477, log1070.845\log_{10}7\approx0.845 covers most interview cases via products of small primes.

Before reaching for logs, check whether the exponents share a common factor, dividing both down to smaller exponents with the same base structure (as with 5445^{44} vs 4504^{50} above) can make the comparison obvious without any logarithm at all.

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Practice in interviews

Further reading

  • Engel, Problem-Solving Strategies, ch. 1
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