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Model growth and decay with a factor per period, and use logarithms to evaluate, apply the log rules and solve for an exponent.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
In this lesson you build models of the form $y = a b^{t}$ and read them out of real situations: compound interest, a half-life, a colony that triples, a machine that loses a fixed percentage of its value every year. The one habit that matters is counting periods before anything else. Then you meet the logarithm, which is nothing more than an exponent read backwards, and use it to undo those models — evaluating logs by inspection, rewriting products and powers with the log rules, and finally bringing an unknown exponent down to ground level where ordinary algebra can solve for it.
You already use exponent rules: $2^3 \times 2^4 = 2^7$, $(2^3)^4 = 2^{12}$, $2^{-1} = \frac{1}{2}$. You already turn a percentage into a multiplier, so a $20\%$ rise multiplies by $1.2$ and a $20\%$ fall multiplies by $0.8$. And you know the difference between a linear pattern, which adds the same amount each step, and one that multiplies by the same amount each step. This lesson is that second kind, written down properly.
Growth factor: what you multiply by in one period. A $6\%$ rise has factor $1.06$; a $6\%$ fall has factor $0.94$.
Initial value $a$ in $y = a b^t$: the amount when $t = 0$.
Doubling time and half-life: how long one doubling, or one halving, takes.
Period: the length of time one factor covers. If the factor is per week, $t$ is counted in weeks.
Exponential decay: growth with a factor between $0$ and $1$.
An exponential model is $y = a b^{t}$: a starting amount $a$, multiplied by a fixed growth factor $b$ once per period, with $t$ counting the periods. If $b > 1$ the quantity grows; if $0 < b < 1$ it decays. The whole skill is reading $a$, $b$ and $t$ correctly out of a situation.
Percentage change gives the factor. A $5\%$ annual rise is $b = 1.05$, so $2000$ dollars becomes $2000 \times 1.05^{t}$ after $t$ years: $2205$ after two years, not $2200$, because the second year's interest is charged on the first year's interest too. A $30\%$ annual loss is $b = 0.7$.
Doubling and halving give the period. If a quantity halves every $6$ hours, then after $t$ hours it has been through $\frac{t}{6}$ half-lives, so $y = a \left(\frac{1}{2}\right)^{t/6}$. After $24$ hours that is four halvings: a sixteenth is left. The same shape with $2$ in place of $\frac{1}{2}$ handles doubling time.
The defining property is that equal time intervals multiply by equal factors. A linear model adds $5$ every year; an exponential model multiplies by $1.05$ every year, and over enough years the second overtakes any straight line, whatever its slope.
Another way: table
Two columns of a savings account at $10\%$. Year $0$: $1000$. Year $1$: $1100$. Year $2$: $1210$. Year $3$: $1331$. The differences $100$, $110$, $121$ grow, but the ratios are all exactly $1.1$. Constant ratio, not constant difference, is the signature of an exponential.
Another way: story
A linear quantity walks the same distance each step. An exponential one takes a step that is a fixed fraction of how far it has already gone, so its steps grow with it.
Multiplying by the number of periods instead of raising to it. Three doublings multiply by $2^3 = 8$, not by $2 \times 3 = 6$. Repeated multiplication is what an exponent means.
Mixing up the time unit and the period. If a colony triples every $4$ hours, then after $12$ hours the exponent is $12 \div 4 = 3$, not $12$. Always convert the elapsed time into a count of periods first.
Averaging percentages across years. Losing $10\%$ twice does not lose $20\%$: $0.9 \times 0.9 = 0.81$, so $19\%$ is gone. Percentages of different amounts cannot be added.
Count the periods: $20 \div 5 = 4$.
The exponent counts periods, never hours.
Four triplings multiply by $3^4$.
Once per period, so the factor is repeated four times.
$3^4 = 81$: the colony is $81$ times larger.
Losing $25\%$ leaves $75\%$, so the factor is $0.75$.
Work with what remains, not with what is lost.
Three years: $12000 \times 0.75^3 = 12000 \times 0.421875$.
The factor is applied once per year.
That is $5062.50$. Note it is not $12000 - 75\%$ of anything: each year's loss is smaller than the last.
$12 \div 3 = 4$ half-lives.
$80 \div 2^4 = 80 \div 16 = 5$ g.
A colony multiplies by $3$ every $4$ hours. How many times larger is it after $12$ hours?
The colony is answer times larger.
$112$ g of a substance halves every $6$ days. How much is left after $18$ days?
answer g is left.
A machine bought for $30000$ dollars loses $20\%$ of its value each year. What is it worth after two years?
The machine is worth answer dollars.
A colony multiplies by $4$ every $4$ hours. How many times larger is it after $16$ hours?
The colony is answer times larger.
You have just built models of the form $y = a b^{t}$, and you can compute $y$ from $t$ by raising a number to a power. You already know that subtraction undoes addition and division undoes multiplication, and that inverse operations are how equations get solved. The question this skill answers is the one the last skill left open: given $y$, how do you get $t$ back out of the exponent?
Logarithm: $\log_b y$ is the exponent that turns $b$ into $y$. Read it as "the power of $b$ that gives $y$".
Base $b$: the number being raised. It must be positive and not $1$.
Common logarithm $\log$: base $10$, when no base is written.
Natural logarithm $\ln$: base $e \approx 2.718$.
Argument: the number inside the log. It is always positive — no power of a positive base is ever zero or negative.
A logarithm is an exponent. The two statements $$b^{t} = y \qquad \text{and} \qquad \log_{b} y = t$$ say exactly the same thing, read in opposite directions, so $\log_2 8 = 3$ because $2^3 = 8$, and $\log_{10} 1000 = 3$, and $\log_5 1 = 0$ because anything to the power $0$ is $1$. Converting between the two forms solves most equations on sight: $\log_3 x = 4$ becomes $x = 3^4 = 81$, and $2^{t} = 64$ becomes $t = \log_2 64 = 6$.
Three rules follow from the exponent rules, one for each: $$\log(MN) = \log M + \log N \qquad \log\frac{M}{N} = \log M - \log N \qquad \log(M^{k}) = k \log M$$ Multiplying inside becomes adding outside, because that is what exponents do when powers are multiplied. So $\log(x^3 y) = 3\log x + \log y$, and $\log\frac{x^2}{y} = 2\log x - \log y$.
The third rule is what solves an exponential equation whose answer is not a whole number. From $3^{t} = 50$, take logs of both sides: $t \log 3 = \log 50$, so $t = \frac{\log 50}{\log 3} \approx 3.56$. The exponent has been brought down to ground level, where ordinary algebra can reach it.
Another way: table
Powers of $2$ against their logarithms: $2, 4, 8, 16, 32, 64$ on one line and $1, 2, 3, 4, 5, 6$ underneath. The top row multiplies by $2$ at each step; the bottom row adds $1$. That gap between multiplying and adding is the entire content of the log rules.
Another way: story
A logarithm answers "how many times did you multiply?". It converts a question about size into a question about count, which is why every scale for things that vary enormously — sound, acidity, earthquakes — is a log scale.
"$\log(x + y) = \log x + \log y$." No: it is $\log(xy)$ that equals $\log x + \log y$. A sum inside a log cannot be split at all. Test it: $\log_{10}(10 + 10) \approx 1.3$, while $\log_{10} 10 + \log_{10} 10 = 2$.
Giving the number instead of the exponent. $\log_2 32 = 5$, not $32$. A logarithm is always an exponent, so it is usually a small number even when its argument is enormous.
"$\log$ of a negative is negative." It does not exist. $\log_{10} 0.001 = -3$ is negative — the value of a log may be negative, but the argument never can be.
The question is: what power of $4$ gives $64$?
Say the definition aloud before computing anything.
$4^1 = 4$, $4^2 = 16$, $4^3 = 64$.
Count the factors.
So $\log_4 64 = 3$.
The unknown is in the exponent, so take $\log$ of both sides: $\log(5^{t}) = \log 200$.
Both sides stay equal because $\log$ is a function.
The power rule brings $t$ down: $t \log 5 = \log 200$.
This is the step that makes the equation solvable.
$t = \frac{\log 200}{\log 5} \approx \frac{2.301}{0.699} \approx 3.29$. Check: $5^3 = 125$ and $5^4 = 625$, so $t$ between $3$ and $4$ is right.
The quotient splits into a difference: $\log(x^5) - \log y$.
The power comes down: $5\log x - \log y$.
What is $\log_{3} 9$?
The logarithm is answer.
Solve $\log_{3} x = 3$.
The solution is x = answer.
Writing $L = \log x$ and $M = \log y$, write $\log(x^{2} y)$ in terms of $L$ and $M$.
Answer:
What is $\log_{5} 125$?
The logarithm is answer.
The Richter magnitude of an earthquake is $\log_{10}$ of the amplitude of the shaking it causes. Quake A shook the ground $100$ times as much as quake B. How much higher is A's magnitude than B's?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$32$ g of a substance halves every $6$ days. How much is left after $6$ days?
answer g is left.
Solve $\log_{2} x = 2$.
The solution is x = answer.
You can build and undo an exponential model. Say what $500$ dollars at $8\%$ a year is worth after three years, evaluate $\log_3 81$, and solve $2^{t} = 1000$ to two decimal places.
8. Your turn: $80$ g halves every $3$ days. How much is left after $12$ days?, step 2
19. Your turn: write $\log\frac{x^5}{y}$ in terms of $\log x$ and $\log y$, step 2