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College Algebra

Change of Base

Lesson

Calculators only have log\log (base 10) and ln\ln (base ee). The change-of-base formula rewrites a log in any base as a ratio of logs in a base you can compute.

The formula

logb(x)=logc(x)logc(b)\log_b(x) = \frac{\log_c(x)}{\log_c(b)}

The new base cc can be anything you like — usually 10 (with log\log) or ee (with ln\ln).

Worked example 1 — clean integer answer

log4(64)=log2(64)log2(4)=62=3\log_4(64) = \frac{\log_2(64)}{\log_2(4)} = \frac{6}{2} = 3

We picked base 2 because both 64 and 4 are nice powers of 2.

Worked example 2 — non-integer answer

log4(8)=log2(8)log2(4)=32\log_4(8) = \frac{\log_2(8)}{\log_2(4)} = \frac{3}{2}

Worked example 3 — using natural log

log5(20)=ln20ln52.9961.6091.861\log_5(20) = \frac{\ln 20}{\ln 5} \approx \frac{2.996}{1.609} \approx 1.861

When to use it

  • Both numbers are powers of the same base → use that base, get an exact answer.
  • Numbers aren’t nice powers → use ln\ln or log\log for a decimal.

Practice

Work through these. Stuck? Click Get a hint.

Warm-Up

Quick problems to get going.

Problem 1

log2(8)\log_2(8)

Problem 2

log3(27)\log_3(27)

Problem 3

log5(125)\log_5(125)

Problem 4

log2(16)\log_2(16)

Practice

Standard problems matching the lesson.

Problem 5

log2(32)\log_2(32)

Problem 6

log3(81)\log_3(81)

Problem 7

log4(64)\log_4(64)

Problem 8

log5(625)\log_5(625)

Problem 9

log10(1000)\log_{10}(1000)

Problem 10

log2(1)\log_2(1)

Problem 11

log7(49)\log_7(49)

Problem 12

log3(243)\log_3(243)

Problem 13

log2(64)\log_2(64)

Problem 14

log8(512)\log_8(512)

Problem 15

log9(81)\log_9(81)

Problem 16

log2(2)\log_2(2)

Problem 17

Signal halves each km; how many km to 1/8?

Problem 18

log5(1)\log_5(1)

Challenge

Harder problems — edge cases, trickier numbers, multiple steps.

Problem 19

log16(64)\log_{16}(64)

Problem 20

log4(8)\log_4(8)

Problem 21

log27(9)\log_{27}(9)

Problem 22

log25(125)\log_{25}(125)

Problem 23

log8(16)\log_8(16)

Problem 24

log4(2)\log_4(2)

Problem 25

log9(3)\log_9(3)

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