2.2, 3.2 - Logarithms and their laws

2.2, 3.2 - Logarithms and their laws

A logarithm answers a power question: which exponent produces a given number? In this shared part of 2.2 and 3.2, you will translate between logarithms and indices, combine logarithms using their laws, and solve equations while keeping every logarithm's input positive.

Logarithms undo powers

What power of 22 gives 88, and what power gives 1/81/8? The first answer is 33. Since 2−3=1/23=1/82^{-3}=1/2^3=1/8, the second is −3-3. A logarithm records exactly these answers.

Logarithm

For a base a>0a>0, a≠1a\ne1, and an argument u>0u>0,

log⁡au=v⟺av=u.\log_a u=v\quad\Longleftrightarrow\quad a^v=u.

The argument is the number whose logarithm is taken. The logarithm is an exponent, so it may be negative, zero or positive.

The restrictions matter. Positive-base powers are positive, so real logarithms of zero or negative numbers are undefined. Base 11 cannot distinguish exponents because 1v=11^v=1. Negative bases do not define real powers for every real exponent. A base between 00 and 11 is allowed: for example, log⁡1/28=−3\log_{1/2}8=-3.

Worked example 1: a positive logarithm

Question. Evaluate log⁡28\log_2 8.

Plan. Write the argument as a power of the base.

8=23⟹log⁡28=3.8=2^3\quad\Longrightarrow\quad\log_2 8=3.

Answer. 33.

Check. Raising the base to the answer gives 23=82^3=8, the original argument.

Replace the argument by its reciprocal. The exponent changes sign; the logarithm remains defined because the argument is still positive.

Worked example 2: a negative logarithm

Question. Evaluate log⁡2(1/8)\log_2(1/8).

Plan. Use a negative power to write the reciprocal in base 22.

18=2−3⟹log⁡2(18)=−3.\frac18=2^{-3}\quad\Longrightarrow\quad\log_2\left(\frac18\right)=-3.

Answer. −3-3.

Check. 2−3=1/82^{-3}=1/8; a negative logarithm has produced a positive number when used as an exponent.

Because a0=1a^0=1 and a1=aa^1=a, every permitted base satisfies log⁡a1=0\log_a1=0 and log⁡aa=1\log_aa=1. Fractional answers work too: log⁡93=1/2\log_93=1/2 since 91/2=39^{1/2}=3.

Watch out

log⁡2(1/8)=−3\log_2(1/8)=-3 is valid. log⁡2(−8)\log_2(-8) is undefined over the reals. A negative answer and a negative argument are different things.

Translate a new logarithm into an index equation.

Combining logarithms

Multiplying powers of the same base adds their exponents: apaq=ap+qa^p a^q=a^{p+q}. Logarithms reverse powers, so a product becomes a sum of logarithms. Division subtracts exponents, and raising a power multiplies exponents.

Laws of logarithms

For a>0a>0, a≠1a\ne1, positive u,vu,v, and real rr,

log⁡a(uv)=log⁡au+log⁡av,\log_a(uv)=\log_a u+\log_a v, log⁡a(uv)=log⁡au−log⁡av,log⁡a(ur)=rlog⁡au.\log_a\left(\frac uv\right)=\log_a u-\log_a v, \qquad \log_a(u^r)=r\log_a u.

All logarithms in a combination use the same base. Not in the formula booklet: learn them.

For instance, if u=apu=a^p and v=aqv=a^q, then uv=ap+quv=a^{p+q}; taking its logarithm gives p+qp+q, proving the product law. The quotient and power laws follow from ap/aq=ap−qa^p/a^q=a^{p-q} and (ap)r=apr(a^p)^r=a^{pr}.

Worked example 3: a square inside a logarithm

Question. Evaluate 2log⁡26−log⁡292\log_2 6-\log_2 9 without finding the separate logarithms numerically.

Plan. Bring the coefficient inside as a power, then combine the difference as a quotient.

2log⁡26−log⁡29=log⁡2(62)−log⁡29power law=log⁡2(36/9)quotient law=log⁡24=2\begin{aligned} 2\log_2 6-\log_2 9 &=\log_2(6^2)-\log_2 9 &&\text{power law}\\ &=\log_2(36/9) &&\text{quotient law}\\ &=\log_2 4=2 \end{aligned}

Answer. 22.

Check. The argument after combining is 4=224=2^2. Equivalently, 62=9×226^2=9\times2^2, matching the original exponent relation.

A fractional coefficient instead becomes a root. The same laws still apply because the arguments are positive.

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