Topic 2.12

Logarithmic Function Manipulation

In 2.9 we found that Nova reaches 10,000 subscribers at $t = \log_{1.5} 39.0625$ , somewhere between 9 and 10, but no calculator has $a \log_{1.5}$ key. This note gives the rules that let you rearrange logarithms. One of them, the change of base formula, finally lets you compute that number.

11 MIN READ8 IDEAS34 PROBLEMS14 flashcards

Practice 34 problems Next: Chapter 2.13A Exponential and Logarithmic Equations and Inequalities

Remember from 2.4: the exponent rules $b^{m} \cdot b^{n} = b^{m+n}$ , $b^{m} / b^{n} = b^{m-n}$ , and $(b^{m})^{p} = b^{mp}$ . A logarithm IS an exponent (2.9), so every exponent rule turns into a logarithm rule. If you ever forget a log rule, rebuild it from the exponent rule, as we do below.

A note on notation: $\log_{b}$ means “logarithm with base b” (the $b$ is written small and low in print). log without a base means base 10, and ln means base $e$ .

The Three Rules

CONCEPT

Properties of Logarithms ( $M$ , $N > 0$ )

Product rule: the log of a product is the SUM of the logs.

Quotient rule: the log of a quotient is the DIFFERENCE of the logs.

Power rule: an exponent inside the log comes out in front as a MULTIPLIER.

\log_b(MN) = \log_b M + \log_b N \qquad \log_b\left(\tfrac{M}{N}\right) = \log_b M - \log_b N \qquad \log_b(M^{p}) = p\log_b M

Each rule comes straight from an exponent rule. Name the two logarithms first: let $m = \log_{b} M$ and $n = \log_{b} N$ . By the definition of a logarithm, that means $M = b^{m}$ and $N = b^{n}$ .

01
Product: multiplying powers ADDS exponents, so the log (the exponent) of $MN$ is $m + n$ .

M = b^{m},\; N = b^{n} \;\Rightarrow\; MN = b^{m+n} \;\Rightarrow\; \log_b(MN) = m + n = \log_b M + \log_b N

02
Quotient: dividing powers SUBTRACTS exponents.

\frac{M}{N} = \frac{b^{m}}{b^{n}} = b^{m-n} \;\Rightarrow\; \log_b \frac{M}{N} = m - n = \log_b M - \log_b N

03
Power: raising a power to a power MULTIPLIES exponents.

M^{p} = (b^{m})^{p} = b^{mp} \;\Rightarrow\; \log_b(M^{p}) = mp = p\log_b M

Now check each rule with numbers you can do by hand:

rule	left side	right side
product	$\log_{2}(8 \cdot 4) = \log_{2} 32 = 5$	$\log_{2} 8 + \log_{2} 4 = 3 + 2 = 5$
quotient	$\log_{3}(81/9) = \log_{3} 9 = 2$	$\log_{3} 81 - \log_{3} 9 = 4 - 2 = 2$
power	$\log_{2}(8^{2}) = \log_{2} 64 = 6$	$2 \cdot \log_{2} 8 = 2 \cdot 3 = 6$

Using the Rules With Known Values

The rules let you build new logarithms from ones you already know. Suppose you are told $\log 2 \approx 0.3010$ and $\log 3 \approx 0.4771$ .

Worked example

Use $\log 2 \approx 0.3010$ and $\log 3 \approx 0.4771$ to find $\log 6$ , $\log 18$ , $\log(1/9)$ , and $\log 1.5$ .

01
$6 = 2 \cdot 3$ , so by the product rule $\log 6 = \log 2 + \log 3 \approx 0.3010 + 0.4771 = 0.7781$ .
02
$18 = 2 \cdot 3^{2}$ , so $\log 18 = \log 2 + 2 \log 3 \approx 0.3010 + 0.9542 = 1.2552$ .
03
$1/9 = 3^{-2}$ , so by the power rule $\log(1/9) = -2 \log 3 \approx -0.9542$ .
04
$1.5 = 3/2$ , so by the quotient rule $\log 1.5 = \log 3 - \log 2 \approx 0.1761$ . (That's the number you'll see as Nova's slope in 2.15.)
A calculator gives $\log 6 \approx 0.7782$ and $\log 18 \approx 1.2553$ . The last digit differs because $\log 2$ and $\log 3$ were already rounded; rounding errors add up. Keep full calculator values until the final answer.

Expanding a Logarithm

Expanding means breaking one logarithm of a complicated expression into many simple logarithms. Work from the outside in: first split the division, then the multiplication, then bring down the exponents. Rewrite roots as fractional powers first ( $\sqrt{x} = x^{1/2}$ ).

Worked example

Example 1. Expand $\log(100x^{3}/y)$ .

01
Quotient rule: $\log(100x^{3}) - \log y$ .
02
Product rule: $\log 100 + \log x^{3} - \log y$ .
03
Power rule and $\log 100 = 2$ (because $10^{2} = 100$ ):
$\log\left(\frac{100\,x^{3}}{y}\right) = \log 100 + \log x^{3} - \log y = 2 + 3\log x - \log y$

Worked example

Example 2. Expand $\ln(\sqrt{x} / e^{2})$ .

01
Rewrite the root: $\sqrt{x} = x^{1/2}$ . Then the quotient rule: $\ln x^{1/2} - \ln e^{2}$ .
02
Power rule on both, and $\ln e = 1$ :
$\ln\frac{\sqrt{x}}{e^{2}} = \ln x^{1/2} - \ln e^{2} = \tfrac{1}{2}\ln x - 2$

Worked example

Example 3. Expand $\log_{2}(8x^{2} / \sqrt{y})$ .

01
Split into three pieces: $\log_{2} 8 + \log_{2} x^{2} - \log_{2} y^{1/2}$ .
02
$\log_{2} 8 = 3$ , and bring down both exponents:
$\log_2 \frac{8x^{2}}{\sqrt{y}} = \log_2 8 + \log_2 x^{2} - \log_2 y^{1/2} = 3 + 2\log_2 x - \tfrac{1}{2}\log_2 y$

Quick check

Expand $\log_2(8x^3)$ .

Condensing Into One Logarithm

Condensing runs the rules backwards. Order matters: FIRST move every coefficient back inside as an exponent (power rule), THEN combine: plus signs become multiplication, minus signs become division.

Worked example

Example 4. Condense $2 \ln x + \ln 5 - \ln 3$ .

01
Power rule first: $2 \ln x = \ln x^{2}$ .
02
Then combine: the + terms go on top, the − term goes on the bottom:
$2\ln x + \ln 5 - \ln 3 = \ln x^{2} + \ln 5 - \ln 3 = \ln\left(\frac{5x^{2}}{3}\right)$
03
Check at $x = 1$ : the left side is $0 + \ln 5 - \ln 3 \approx 0.5108$ , and the right side is $\ln(5/3) \approx 0.5108$ ✓.

Worked example

Example 5. Condense $\tfrac{1}{2} \log x - 3 \log y + \log 5$ .

01
Power rule on each coefficient: $\tfrac{1}{2} \log x = \log x^{1/2} = \log \sqrt{x}$ , and $3 \log y = \log y^{3}$ .

Combine: $\sqrt{x}$ and 5 are added (top), $y^{3}$ is subtracted (bottom):

\tfrac{1}{2}\log x - 3\log y + \log 5 = \log x^{1/2} - \log y^{3} + \log 5 = \log \frac{5\sqrt{x}}{y^{3}}

Condensing can also simplify numbers: $\ln 12 - 2 \ln 2 = \ln 12 - \ln 4 = \ln(12/4) = \ln 3$ .

The rules only work in these exact shapes. These look similar but are all FALSE. Each row shows a value where the two sides differ:

false statement	left side	right side
$\log(M + N) = \log M$ + $\log N$	$\log(2 + 8) = 1$	$\log 2$ + $\log 8 \approx 1.2041$
$\log(M / N)$ = $\log M$ / $\log N$	$\log(100/10) = 1$	$\log 100$ / $\log 10$ = 2
$(\log M)^{2} = 2 \log M$	$(\log 1000)^{2} = 9$	$2 \log 1000$ = 6
$\log(3M) = 3 \log M$	$\log 30 \approx 1.4771$	$3 \log 10$ = 3

COMMON MISTAKE

“ $\ln(e + e) = \ln e + \ln e = 2$ .”

There is no rule for the log of a SUM. $\ln(e + e) = \ln(2e) = \ln 2 + 1 \approx 1.6931$ . The product rule turns a product inside the log into a sum outside, never the other way around.

“ $(\log x)^{2} = 2 \log x$ by the power rule.”

The power rule needs the exponent INSIDE the log: $\log(x^{2}) = 2 \log x$ . In $(\log x)^{2}$ the whole log is squared, which is a different number (9 versus 6 at $x = 1000$ ).

“ $2 \ln x + \ln 5 = \ln(2x \cdot 5)$ .”

The 2 is an exponent in disguise, not a factor. Move it inside first: $\ln(x^{2} \cdot 5) = \ln(5x^{2})$ .

Worked example

Try it yourself. (a) Expand $\log_{2}(8x^{5})$ . (b) Condense $3 \ln 2 + \ln x - \ln 4$ .

Answers: (a) $3 + 5 \log_{2} x$ . (b) $\ln 8 + \ln x - \ln 4 = \ln(8x/4) = \ln(2x)$ .

Quick check

Write $2\log_5 x - \log_5 y$ as a single logarithm.

Change of Base

CONCEPT

Change of Base Formula

Any logarithm can be computed with the log or ln key: divide the log of the input by the log of the base.

It works with ln or with log, as long as you use the same one on top and bottom.

\log_b c = \frac{\log c}{\log b} = \frac{\ln c}{\ln b}

Why it works: name the unknown logarithm a, rewrite in exponential form, and take ln of both sides:

a = \log_b c \;\Rightarrow\; b^{a} = c \;\Rightarrow\; \ln(b^{a}) = \ln c \;\Rightarrow\; a\ln b = \ln c \;\Rightarrow\; a = \frac{\ln c}{\ln b}

Worked example

Example 6. Compute $\log_{2} 20$ and $\log_{6} 50$ .

01
$\log_{2} 20$ :
$\log_2 20 = \frac{\ln 20}{\ln 2} \approx \frac{2.9957}{0.6931} \approx 4.3219$
02
Check: $2^{4} = 16$ and $2^{5} = 32$ , so an answer between 4 and 5 makes sense (2.9).
03
$\log_{6} 50 = \ln 50 / \ln 6 \approx 2.1833$ . Check: $6^{2.1833} \approx 50.0$ ✓.

REAL-LIFE EXAMPLE

Nova, Solved

Finally: the month when Nova reaches 10,000 subscribers.

\log_{1.5} 39.0625 = \frac{\ln 39.0625}{\ln 1.5} \approx 9.0394

Using log instead of ln gives the same 9.0394. So Nova crosses 10,000 about 9.04 months after launch.

COMMON MISTAKE

“ $\log_{1.5} 39.0625 = \log 39.0625 - \log 1.5$ .”

Change of base is a DIVISION of two logs, not a subtraction. Subtraction belongs to the log of a quotient, $\log(M/N)$ , which is a different expression.

What the Rules Say About Graphs

Each rule is also a statement about transformations of the graph.

CONCEPT

Graphical Meaning

Product rule: $\log_{b}(kx) = \log_{b} k$ + $\log_{b} x$ . Multiplying the input by $k$ (a horizontal stretch or compression) is the same as a vertical SHIFT by $\log_{b} k$ .

Power rule: $\log_{b}(x^{k}) = k \log_{b} x$ for $x > 0$ . Raising the input to a power is a vertical STRETCH by $k$ .

Change of base: $\log_{b} x = \dfrac{1}{\ln b} \cdot \ln x$ . Every logarithmic function is a vertical stretch of $\ln x$ . For example $\log_{2} x \approx 1.4427 \ln x$ and $\log_{3} x \approx 0.9102 \ln x$ .

Multiplying the input by 4 lifts the whole graph of $\log_{2} x$ by $\log_{2} 4 = 2$ .

Two more quick examples: $\log_{2}(8x) = 3 + \log_{2} x$ is the graph of $\log_{2} x$ moved up 3, and $\log(x^{3}) = 3 \log x$ is the graph of $\log x$ stretched vertically by 3 (at $x = 5$ both sides equal $\log 125$ ).

Remember from 2.4: for exponentials, a horizontal SHIFT was the same as a vertical STRETCH. Logarithms swap the roles: a horizontal STRETCH is the same as a vertical SHIFT.

Practice

Worked example

P1. Evaluate $\log_{6} 4 + \log_{6} 9$ without a calculator.

Answer: Product rule: $\log_{6} 36 = 2$ .

Worked example

P2. Evaluate $\log_{3} 54 - \log_{3} 2$ without a calculator.

Answer: Quotient rule: $\log_{3} 27 = 3$ .

Worked example

P3. Use change of base to find $\log_{7} 100$ and $\log_{5} 40$ , to 4 decimal places.

Answer: $\ln 100 / \ln 7 \approx 2.3666$ and $\ln 40 / \ln 5 \approx 2.2920$ .

Worked example

P4. Evaluate $\log_{1/4} 8$ by writing both numbers as powers of 2.

Answer: $(1/4)^{a} = 8$ means $2^{-2a} = 2^{3}$ , so $a = -3/2$ .

Worked example

P5. Expand $\log_{3}(9/x^{4})$ .

Answer: $\log_{3} 9 - \log_{3} x^{4} = 2 - 4 \log_{3} x$ .

Worked example

P6. Condense $2 \log x + \log 3 - \log 12$ .

Answer: $\log x^{2} + \log 3 - \log 12 = \log(3x^{2}/12) = \log(x^{2}/4)$ .

Common slips

“ $\ln(e + e) = \ln e + \ln e = 2$ .”

There is no rule for the log of a sum. $\ln(e + e) = \ln(2e) = \ln 2 + 1 \approx 1.6931$ . The product rule turns a product inside the log into a sum outside, never the other way around.

“ $(\log x)^{2} = 2 \log x$ by the power rule.”

The power rule needs the exponent inside the log: $\log(x^{2}) = 2 \log x$ . In $(\log x)^{2}$ the whole log is squared, which is a different number (9 versus 6 at $x = 1000$ ).

“ $2 \ln x + \ln 5 = \ln(2x \cdot 5)$ .”

The 2 is an exponent in disguise, not a factor. Move it inside first: $\ln(x^{2} \cdot 5) = \ln(5x^{2})$ .
“ $\log_{1.5} 39.0625 = \log 39.0625 - \log 1.5$ .”

Change of base is a division of two logs, not a subtraction. Subtraction belongs to the log of a quotient, $\log(M/N)$ , which is a different expression.

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14 cards · Logarithms, Log rules and equations

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Recap card

5 lines to re-read the night before.

01
Product: $\log(MN) = \log M + \log N$ . Quotient: $\log(M/N) = \log M - \log N$ . Power: $\log(M^{p}) = p \log M$ . Each comes from an exponent rule.
02
Expand from the outside in (division, then multiplication, then exponents). Condense in the reverse order: coefficients go inside first.
03
There is no rule for $\log(M + N)$ , $\log M / \log N$ , or $(\log M)^{2}$ ; test suspicious steps with numbers.
04
Change of base: $\log_{b} c = \ln c / \ln b$ . This computes any logarithm, like Nova's $\log_{1.5} 39.0625 \approx 9.0394$ .
05
Graphs: $\log(kx)$ is a vertical shift of $\log x$ ; $\log(x^{k})$ is a vertical stretch. Next, in 2.13: solving exponential and logarithmic equations.

Practice 34 problems Next: Chapter 2.13A Exponential and Logarithmic Equations and Inequalities

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Logarithmic Function Manipulation

The Three Rules

Using the Rules With Known Values

Expanding a Logarithm

Condensing Into One Logarithm

Change of Base

What the Rules Say About Graphs

Practice

Recap card

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