1B - Combining logarithms

Can adding two logarithms turn a difficult calculation into an exact power? In 1B you will use the index laws to explain the logarithm laws, then expand and combine expressions. The same-base and positive-argument conditions will keep every rearrangement valid.

Products and quotients

Adding exponents multiplies powers, so adding logarithms combines their arguments by multiplication.

For example, $2^3\times2^4=2^7$ . Since $\log_2 8=3$ and $\log_2 16=4$ , we have $\log_2(8\times16)=3+4$ . The logarithm records the exponent, so the addition comes from the index law.

To see why this always works, write $u=\log_a x$ and $v=\log_a y$ . Then $x=a^u$ , $y=a^v$ , and

xy=a^{u+v},\qquad \frac{x}{y}=a^{u-v}.

Taking the logarithm to base $a$ reads off those exponents.

Product and quotient laws

For a fixed base $a>1$ and positive $x$ and $y$ ,

\log_a(xy)=\log_a x+\log_a y, \qquad \log_a\!\left(\frac{x}{y}\right)=\log_a x-\log_a y.

The logarithms being combined must have the same base. These laws must be learnt; they are not on the examination formulae sheet.

You may use either direction: a product can be split into a sum, or a sum can be condensed into one logarithm. A subtraction keeps its order: the first argument goes in the numerator.

Worked example 1: combining a sum

Question. Find the exact value of $\log_5 2+\log_5(25/2)$ .

Plan. Combine the same-base logarithms into a product whose value is a power of $5$ .

\begin{aligned} \log_5 2+\log_5\!\left(\frac{25}{2}\right) &=\log_5\!\left(2\times\frac{25}{2}\right) &&\text{product law}\\ &=\log_5 25=2 &&\text{since }5^2=25 \end{aligned}

Answer. $2$ .

Check. Numerically, the original logarithms are approximately $0.430677$ and $1.569323$ ; their sum is $2$ .

Now change addition to subtraction. Multiplication of the arguments must change to division.

Worked example 2: combining a difference

Question. Find the exact value of $\log_5 250-\log_5 2$ .

Plan. Divide the first argument by the second.

\begin{aligned} \log_5 250-\log_5 2 &=\log_5\!\left(\frac{250}{2}\right) &&\text{quotient law}\\ &=\log_5 125=3 &&\text{since }5^3=125 \end{aligned}

Answer. $3$ .

Check. The original values are approximately $3.430677$ and $0.430677$ , with difference $3$ .

The laws combine whole logarithm values. They do not say that $\log_5 250-\log_5 2=\log_5 248$ , and they cannot directly combine logarithms with different bases.

Powers and reciprocals

A power inside a logarithm becomes a multiplier outside it.

If $u=\log_a x$ , then $x=a^u$ . Raising both sides to a power $k$ gives $x^k=a^{ku}$ , so the logarithm of $x^k$ is $ku$ .

Power and reciprocal laws

For $a>1$ , $x>0$ and real $k$ ,

\log_a(x^k)=k\log_a x.

In particular, $x^{-1}=1/x$ gives

\log_a\!\left(\frac1x\right)=-\log_a x.

These laws must be learnt; they are not on the examination formulae sheet.

The condition $x>0$ matters even when the power is a square: $\log_2((-3)^2)$ exists, but $2\log_2(-3)$ does not. You may not expand the first expression that way.

Worked example 3: taking a positive power outside

Question. Find the exact value of $\log_2(8^2)$ .

Plan. Use the power law, then evaluate $\log_2 8$ .

\log_2(8^2)=2\log_2 8=2\times3=6.

The exponent $2$ multiplies the logarithm; it does not multiply the argument $8$ .

Answer. $6$ .

Check. The argument is $64$ , and $2^6=64$ .

Keep the argument built from $8$ , but use a reciprocal square root. Its exponent is negative and fractional.

Worked example 4: a reciprocal root

Question. Find the exact value of $\log_2\!\left(1/\sqrt8\right)$ .

Plan. Express the argument as $8^{-1/2}$ , then use the power law.

\begin{aligned} \log_2\!\left(\frac1{\sqrt8}\right) &=\log_2(8^{-1/2}) &&\text{write a negative fractional power}\\ &=-\frac12\log_2 8=-\frac32 &&\text{take the exponent outside} \end{aligned}

Answer. $-\frac32$ .

Check. $2^{-3/2}=1/\sqrt{2^3}=1/\sqrt8$ , which recovers the original argument.

There is no corresponding law that splits a sum of arguments. For instance,

\log_2(4+4)=\log_2 8=3, \qquad \log_2 4+\log_2 4=2+2=4.

The second expression is the logarithm of $4\times4$ , not of $4+4$ . Look inside the brackets before choosing a law.

Equivalent logarithmic forms

Several logarithm laws can be used together, but an equivalent form must keep the original domain.

To expand a logarithm, separate products and quotients, then bring powers outside. To condense, first turn multipliers into powers, then combine sums and differences. Brackets show which whole expression is the argument or the exponent.

Worked example 5: expanding with valid arguments

Question. Expand $\log_3\!\left(\frac{9(x+1)^2}{x-2}\right)$ , stating its domain.

Plan. Find where the original argument is positive before splitting it.

The numerator is positive except at $x=-1$ , where it is zero. A positive quotient requires $x-2>0$ , so the original domain is $x>2$ . On this domain both $x+1$ and $x-2$ are positive.

\begin{aligned} \log_3\!\left(\frac{9(x+1)^2}{x-2}\right) &=\log_3 9+\log_3((x+1)^2)-\log_3(x-2)\\ &=2+2\log_3(x+1)-\log_3(x-2) \end{aligned}

Answer. $2+2\log_3(x+1)-\log_3(x-2)$ , for $x>2$ .

Check. At $x=3$ , the original is $\log_3 144$ . The expanded form is $2+2\log_3 4$ , whose exponential value is $3^2\times4^2=144$ .

Now work in the opposite direction. Keeping the restrictions is especially important because a condensed argument can be positive even when an original argument is negative.

Worked example 6: condensing without extending the domain

Question. Write $\log_2(x-1)+\log_2(x+2)$ as one logarithm, retaining its domain.

Plan. Require each original argument to be positive, then use the product law.

x-1>0\quad\text{and}\quad x+2>0 \quad\Longrightarrow\quad x>1.

Hence

\log_2(x-1)+\log_2(x+2)=\log_2((x-1)(x+2)),\qquad x>1.

Answer. $\log_2((x-1)(x+2))$ , with $x>1$ .

Check. At $x=2$ , both forms equal $2$ . At $x=-3$ , the condensed argument would be $4$ , but the original logarithms have arguments $-4$ and $-1$ . That extra input must stay excluded.

The laws establish equality wherever the original steps are defined. They do not grant extra permitted inputs to a rearranged expression.

Key idea

Use the same base, attach each sign to its whole term and retain every original positive-argument condition. A sum inside one logarithm does not split into a sum of logarithms.

Apply this in the reverse direction to powers and a quotient.

Quick Recap

Adding logarithms multiplies their arguments; subtracting divides them; a multiplier becomes a power. Each rule comes from an index law. The opening calculation becomes exact when the combined argument is a known power, as in $\log_5 2+\log_5(25/2)=2$ .

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1B - Combining logarithms

Products and quotients

Powers and reciprocals

Equivalent logarithmic forms

Quick Recap

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