Unit 3, lesson 41

The Power Rule and Change of Base for Logarithms

How an exponent inside a logarithm moves out in front as a factor, why that needs a positive argument, and how to rewrite any logarithm in base 10 or base e, exactly or as a decimal.

  • Theory part 81
  • Practice part 82
  • Properties of Logarithms

How many times must you double 1 to pass a million? The exact answer to 2x=1062^x = 10^6 is x=log⁡2(106)x = \log_2\left(10^6\right), yet most calculators have keys only for log⁡\log and ln⁡\ln. Two rules close the gap. The power rule moves an exponent out of a logarithm, and the change of base formula rewrites a logarithm in a base your calculator knows. Together they give x=6/log⁡2≈19.93x = 6/\log 2 \approx 19.93, so it takes 20 doublings.

Key terms

The power rule

Definition Let b>0b > 0 with b≠1b \ne 1, let M>0M > 0, and let rr be any real number. Then

log⁡b(Mr)=rlog⁡bM.\log_b\left(M^r\right) = r \log_b M.

To prove it, put u=log⁡bMu = \log_b M, so that bu=Mb^u = M. Then Mr=(bu)r=burM^r = \left(b^u\right)^r = b^{ur}, and by the definition of the logarithm log⁡b(Mr)=ur=rlog⁡bM\log_b\left(M^r\right) = ur = r \log_b M. Roots are powers too, so for every integer n≥2n \ge 2 the rule gives

log⁡bMn=1nlog⁡bM.\log_b \sqrt[n]{M} = \frac{1}{n} \log_b M.

In plain words An exponent inside a logarithm can come out in front as a multiplier. This is the step that brings an unknown down from an exponent when you solve equations.

Example Since 8=238 = 2^3,

log⁡2(85)=5log⁡28=5⋅3=15,\log_2\left(8^5\right) = 5 \log_2 8 = 5 \cdot 3 = 15,

and indeed 85=32768=2158^5 = 32768 = 2^{15}. The rule also counts digits of numbers far too long to write out by hand. The common logarithm of 21002^{100} is

log⁡(2100)=100log⁡2≈30.103,\log\left(2^{100}\right) = 100 \log 2 \approx 30.103,

so the number lies between 103010^{30} and 103110^{31} and has 31 digits.

Positive argument in the power rule

Definition The power rule needs M>0M > 0. The left side log⁡b(Mr)\log_b\left(M^r\right) can be defined for a negative MM, for example when rr is an even integer, but then log⁡bM\log_b M is not defined. For every real M≠0M \ne 0 and every integer kk the correct statement is

log⁡b(M2k)=2klog⁡b∣M∣.\log_b\left(M^{2k}\right) = 2k \log_b \lvert M \rvert.

So ln⁡(x2)=2ln⁡∣x∣\ln\left(x^2\right) = 2 \ln \lvert x \rvert for every x≠0x \ne 0, while ln⁡(x2)=2ln⁡x\ln\left(x^2\right) = 2 \ln x holds only for x>0x > 0.

In plain words Before you pull an exponent out, make sure the number being raised to it is positive. If it might be negative, put absolute value bars around it.

Example At M=−3M = -3 the left side is ln⁡((−3)2)=ln⁡9≈2.1972\ln\big((-3)^2\big) = \ln 9 \approx 2.1972, but 2ln⁡(−3)2 \ln(-3) is undefined. The version with absolute values gives 2ln⁡3≈2.19722 \ln 3 \approx 2.1972, which matches. The domains differ as well. The expression ln⁡(x2)\ln\left(x^2\right) is defined for every x≠0x \ne 0, while 2ln⁡x2 \ln x is defined only for x>0x > 0, so trading one for the other in an equation can lose solutions or add false ones.

Change of base

Definition Let aa and bb be positive numbers, neither equal to 11, and let x>0x > 0. Then

log⁡bx=log⁡axlog⁡ab.\log_b x = \frac{\log_a x}{\log_a b}.

To prove it, put y=log⁡bxy = \log_b x, so that by=xb^y = x. Taking log⁡a\log_a of both sides and using the power rule gives ylog⁡ab=log⁡axy \log_a b = \log_a x. Since b≠1b \ne 1, the number log⁡ab\log_a b is not 00, and dividing by it gives the formula. With the two bases on a calculator,

log⁡bx=log⁡xlog⁡b=ln⁡xln⁡b.\log_b x = \frac{\log x}{\log b} = \frac{\ln x}{\ln b}.

Taking x=ax = a gives the special case log⁡ba=1/log⁡ab\log_b a = 1/\log_a b. Calculus uses the formula to reduce every logarithm to ln⁡\ln, for example in derivatives of exponential and logarithmic functions. The power rule does the same for exponentials. For b>0b > 0 and every real xx it gives ln⁡(bx)=xln⁡b\ln\left(b^x\right) = x \ln b, so bx=exln⁡bb^x = e^{x \ln b}. Calculus takes this formula as its definition of bxb^x, and the computation shows that it gives the same values as the definition in the lesson on exponential functions.

In plain words Logarithms in two different bases differ only by a constant factor, the way a length in inches and the same length in centimeters do. To switch to a new base, take the logarithm of xx in the new base and divide it by the logarithm of the old base, also taken in the new base.

Example Compare base 2 with base 10 on a few powers of 2.

xxlog⁡2x\log_2 xlog⁡x\log xlog⁡x\log x divided by log⁡2x\log_2 x
22110.301030.301030.301030.30103
44220.602060.602060.301030.30103
88330.903090.903090.301030.30103
1616441.204121.204120.301030.30103

Every row has the same ratio, log⁡2≈0.30103\log 2 \approx 0.30103. That is the change of base formula in the form log⁡x=log⁡2⋅log⁡2x\log x = \log 2 \cdot \log_2 x. Solving for log⁡2x\log_2 x at x=10x = 10 gives

log⁡210=log⁡10log⁡2=ln⁡10ln⁡2≈3.32193.\begin{aligned} \log_2 10 &= \frac{\log 10}{\log 2} \\[0.8em] &= \frac{\ln 10}{\ln 2} \approx 3.32193. \end{aligned}

Valid logarithm bases

Definition The base of a logarithm must be a positive real number other than 11, and in the change of base formula this applies to both bases. The new base aa must be valid so that log⁡ax\log_a x and log⁡ab\log_a b exist, and the condition b≠1b \ne 1 is exactly what keeps the denominator log⁡ab\log_a b away from 00. When the base contains a variable, as in log⁡x4\log_x 4, the expression is defined only for x>0x > 0 with x≠1x \ne 1.

In plain words Any valid base can serve as the new base, and the result does not depend on which one you pick. Bases 10 and ee are popular only because calculators have keys for them.

Example Solve log⁡x4=2\log_x 4 = 2. By the definition of the logarithm this means x2=4x^2 = 4, so x=2x = 2 or x=−2x = -2. A base must be positive, so −2-2 is rejected and the only solution is x=2x = 2. Change of base shows the same restrictions from another side. The formula

log⁡x4=ln⁡4ln⁡x\log_x 4 = \frac{\ln 4}{\ln x}

needs x>0x > 0 for ln⁡x\ln x to exist and fails at x=1x = 1, where the denominator ln⁡1\ln 1 is 00.

Exact expressions

Definition An exact expression names a number with no rounding, using integers, fractions, radicals, constants such as π\pi and ee, and functions such as logarithms. The expressions log⁡350\log_3 50, ln⁡50ln⁡3\frac{\ln 50}{\ln 3} and log⁡32+2log⁡35\log_3 2 + 2 \log_3 5 are all exact, and they all name the same number. An exact answer is usually written in simplest form, with every logarithm that equals a rational number replaced by that number, for example log⁡28\log_2 8 by 33.

In plain words Keep answers exact for as long as you can. An exact answer can always be turned into a decimal later, but a decimal cannot be turned back into the exact answer.

Example When the argument and the base are integer powers of the same number, change of base gives an exact fraction. Since 32=2532 = 2^5 and 4=224 = 2^2,

log⁡432=ln⁡32ln⁡4=5ln⁡22ln⁡2=52.\log_4 32 = \frac{\ln 32}{\ln 4} = \frac{5 \ln 2}{2 \ln 2} = \frac{5}{2}.

The same idea gives log⁡832=53\log_8 32 = \frac{5}{3} and log⁡927=32\log_9 27 = \frac{3}{2}. To check the first value with powers, note that 45/2=(4)5=25=324^{5/2} = \left(\sqrt{4}\right)^5 = 2^5 = 32.

Calculator approximations

Definition A calculator approximation is a decimal that agrees with an exact value to a stated number of decimal places or significant digits. It is written with ≈\approx rather than ==. Rounding in the middle of a calculation lets the errors grow, so keep full precision in every intermediate step and round only the final answer.

In plain words Type the whole exact expression into the calculator in one go, then round once. Raising the base to your answer is a quick check that catches most slips.

Example Entering the whole quotient at once gives

log⁡350=ln⁡50ln⁡3≈3.56088,\log_3 50 = \frac{\ln 50}{\ln 3} \approx 3.56088,

and the check 33.56088≈50.00023^{3.56088} \approx 50.0002 confirms it. Rounding first to ln⁡50≈3.91\ln 50 \approx 3.91 and ln⁡3≈1.10\ln 3 \approx 1.10 gives 3.91÷1.10≈3.554553.91 \div 1.10 \approx 3.55455, which is already off in the second decimal place, and 33.554553^{3.55455} is only about 49.6549.65.

Check yourself

  1. Find log⁡84\log_8 4 exactly.
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Both numbers are powers of 2, so

log⁡84=ln⁡4ln⁡8=2ln⁡23ln⁡2=23.\log_8 4 = \frac{\ln 4}{\ln 8} = \frac{2 \ln 2}{3 \ln 2} = \frac{2}{3}.

Indeed 82/3=(83)2=48^{2/3} = \left(\sqrt[3]{8}\right)^2 = 4.

  1. Approximate log⁡540\log_5 40 to four decimal places and check your answer.
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By change of base

log⁡540=ln⁡40ln⁡5≈2.2920.\log_5 40 = \frac{\ln 40}{\ln 5} \approx 2.2920.

As a check, 52.2920≈39.9985^{2.2920} \approx 39.998, which is 4040 up to the rounding.

  1. Use the power rule to find how many digits the number 3503^{50} has.
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By the power rule

log⁡(350)=50log⁡3≈23.856,\log\left(3^{50}\right) = 50 \log 3 \approx 23.856,

so the number lies between 102310^{23} and 102410^{24} and has 24 digits.

How this lesson connects