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The product, quotient and power rules for logarithms, solving exponential and logarithmic equations with a domain check, and modelling growth, decay, cooling and logarithmic scales.
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 logarithms stop being a definition and become a tool. You learn the three rules that turn a product into a sum and a power into a multiplier, use change of base to compute a logarithm your calculator has no key for, and solve equations by collapsing to one logarithm and rewriting it as a power — always checking the answers, because a logarithm refuses a negative argument and rearranging invents roots. Then you point the same machinery at the world: continuous growth, half-lives, doubling times, Newton's cooling and the logarithmic scales that measure acidity and sound.
You can already solve $2^x = 32$ by recognising $32 = 2^5$, and you know that $\log_b x$ is the exponent you must put on $b$ to get $x$. You can solve a quadratic by factoring. This lesson does two things with those: it gives you the three rules that let you combine logarithms before solving, and it makes you check the answers, because a logarithm refuses some of them.
Logarithm: $\log_b x = k$ means $b^k = x$. The base $b$ is positive and not $1$.
Common logarithm, written $\log x$: base $10$.
Natural logarithm, written $\ln x$: base $e \approx 2.718$.
Argument: the thing inside the logarithm. It must be strictly positive.
Change of base: rewriting $\log_b N$ as a ratio of logarithms in a base your calculator has.
Extraneous root: a value that solves the equation you rearranged to but not the one you were given.
Every rule for logarithms is a rule for exponents read backwards. Product: $\log_b(xy) = \log_b x + \log_b y$. Quotient: $\log_b\frac{x}{y} = \log_b x - \log_b y$. Power: $\log_b x^p = p\log_b x$. Together they let you turn any product of powers into a sum, so $\log(1000x^2) = 3 + 2\log x$. Change of base lets you compute one you have no key for: $\log_b N = \frac{\log N}{\log b}$, so $\log_2 9 = \frac{\log 9}{\log 2}$. To solve, get to one logarithm on one side and then rewrite it as a power: $\log_3 x + \log_3(x-2) = 1$ becomes $x(x-2) = 3^1$, giving $x = 3$ or $x = -1$. Finally, and always: check the domain. Every argument must be strictly positive, so $x = -1$ is thrown away and only $x = 3$ survives. The domain of $y = \log(x - 3)$ is $x > 3$ for the same reason.
Change of base needs the logarithm of a few small numbers, and these are the ones this lesson draws on:
| $n$ | $\log_{10} n$ |
|---|---|
| $2$ | $0.3010$ |
| $3$ | $0.4771$ |
| $5$ | $0.6990$ |
| $6$ | $0.7782$ |
| $7$ | $0.8451$ |
| $11$ | $1.0414$ |
| $12$ | $1.0792$ |
| $15$ | $1.1761$ |
Everything else follows from the rules: $\log_{10} 4 = 2\log_{10} 2$, $\log_{10} 20 = 1 + \log_{10} 2$, and $\log_{10} 5 = 1 - \log_{10} 2$.
Another way: picture
The graph of $y = \log x$: it climbs steeply out of the bottom of the page just to the right of $x = 0$, crosses the axis at $(1, 0)$, and then flattens — never reaching the left of $x = 0$ at all, which is the domain restriction drawn.
Another way: steps
"$\log(a + b) = \log a + \log b$." No. Addition inside has no rule at all; it is multiplication inside that becomes addition outside: $\log(ab) = \log a + \log b$.
"$\frac{\log 8}{\log 2} = \log 4$." No — dividing two logarithms is change of base, not the quotient rule. $\frac{\log 8}{\log 2} = \log_2 8 = 3$, while $\log 4 \approx 0.602$.
"Every root of the rearranged equation is a solution." Combining logarithms can invent roots. $\log_3 x + \log_3(x - 2) = 1$ becomes $x^2 - 2x - 3 = 0$ with roots $3$ and $-1$, and $-1$ is not a solution: $\log_3(-1)$ does not exist.
Product rule: $\log_2\big(x(x-2)\big) = 3$.
One logarithm, alone on the left.
Rewrite as a power: $x(x - 2) = 2^3 = 8$, so $x^2 - 2x - 8 = 0$ and $(x - 4)(x + 2) = 0$.
The roots are $4$ and $-2$; $\log_2(-2)$ does not exist, so the solution is $x = 4$.
Always check the domain.
Change of base with the new base on the bottom: $\log_2 5 = \frac{\log_{10} 5}{\log_{10} 2}$.
$\frac{0.6990}{0.3010} \approx 2.322$. Sanity check: $2^2 = 4$ and $2^3 = 8$, so the answer had to be between $2$ and $3$.
The argument must be positive: $5 - x > 0$.
So $x < 5$: the domain is $(-\infty, 5)$, open at $5$.
Solve $\log_{5} x = 2$.
x = answer.
Solve $4^{\,x + 3} = 65536$.
x = answer.
Solve $\log_{2} x + \log_{2}(x - 15) = 4$.
The admissible solution is x = answer.
Using $\log_{10} 11 = 1.0414$ and $\log_{10} 6 = 0.7782$, find $\log_{11} 6$ to three decimal places.
The logarithm is approximately answer.
You have met percentage growth and compound interest, and you have just learned the three rules for logarithms and how to solve with them. What is new here is not a technique but a habit: reading a real situation, deciding it is exponential, and naming which letter in $Ae^{rt}$ each given number is.
Continuous growth: $A(t) = Pe^{rt}$, with $P$ the starting amount and $r$ the rate per unit time written as a decimal.
Half-life: the time in which half of a decaying quantity is left.
Doubling time: the time in which a growing quantity doubles.
Newton's law of cooling: $T = T_{\text{room}} + (T_0 - T_{\text{room}})e^{-kt}$.
Logarithmic scale: one where equal steps mean equal factors — pH, decibels, the Richter scale.
Semi-log plot: $\log y$ against $x$; an exponential model plots as a straight line.
One shape covers most of it: $A(t) = A_0 e^{rt}$, where $A_0$ is the value at $t = 0$ and $r$ is the continuous rate — positive for growth, negative for decay. A population growing at $5\%$ a year is $A_0e^{0.05t}$, so after $10$ years it is $A_0e^{0.5}$. Half-life is the same idea counted in halvings: $N = N_0\left(\tfrac{1}{2}\right)^{t/h}$, so from $100$ mg to $25$ mg is two halvings and takes $2h$. Doubling time is what you get by solving $e^{rt} = 2$, namely $t = \frac{\ln 2}{r}$, which is why a rate of $0.07$ doubles in about ten years. Cooling adds a floor: $T = 20 + 60e^{-kt}$ starts at $80^\circ$ and settles at $20^\circ$, the temperature of the room. And a logarithmic scale turns multiplication into addition: pH is $-\log_{10}[\mathrm{H}^+]$, so ten times the acid is one unit lower. Whenever data plotted as $\log y$ against $x$ falls on a line, the underlying model is exponential.
Another way: picture
Two axes side by side with the same decay data: on the left, $N$ against $t$ curving down towards but never touching the axis; on the right, $\log N$ against $t$, the same data as a perfectly straight downward line.
Another way: steps
"$5\%$ means $r = 5$." It means $r = 0.05$. A rate in $Pe^{rt}$ is always a decimal, and forgetting that inflates the answer by a factor of $e^{95}$ or so.
"Two half-lives means none is left." Two half-lives leave a quarter. Halving repeatedly never reaches zero, which is exactly why decay is exponential and not linear.
"A bigger pH is more acidic." The other way round: pH is minus a logarithm, so more hydrogen ions give a smaller number. Every logarithmic scale has this trap somewhere — check which way the sign runs before you answer.
From $100$ to $25$ is two halvings: $100 \to 50 \to 25$.
Count halvings rather than reaching for a logarithm.
Each halving takes the half-life, $8$ years, so $t = 16$ years.
Doubling means $e^{0.07t} = 2$, so $0.07t = \ln 2$.
Take the natural logarithm of both sides.
$t = \frac{\ln 2}{0.07} \approx \frac{0.693}{0.07} \approx 9.9$ years. The doubling time does not depend on where you started.
The starting amount is $A_0 = 500$ and the rate as a decimal is $r = 0.08$.
So $A(t) = 500e^{0.08t}$, and after $10$ hours that is $500e^{0.8}$.
An investment grows continuously at $5\%$ a year, so its value is $Pe^{rt}$. After $5$ years the value is $Pe^{k}$. What is $k$?
The exponent k is answer.
A sample of $144$ mg decays with a half-life of $5$ years. After how many years is $9$ mg left?
The elapsed time is answer years.
A drink cools by Newton's law: $T = 29 + 48e^{-kt}$ degrees after $t$ minutes. What was its temperature when it was poured?
The initial temperature was answer degrees.
A solution has $[\mathrm{H}^+] = 10^{-9}$, so its pH is $9$. A second solution is $100$ times more acidic. What is its pH?
The second solution has pH answer.
Sound level in decibels is $L = 10\log_{10}(I/I_0)$. One machine measures $60$ dB. A second identical machine is switched on beside it, doubling the intensity $I$. What does the meter now read?
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Give the domain of $y = \log(x - 4)$ as an interval.
This task has no paper form; do it on a device.
A biologist plots $\log_{10} y$ against $x$ and the points fall on a straight line. What kind of model does $y$ follow?
You can use the logarithm rules to solve equations and build exponential models. Without looking: which of $\log(a+b)$ and $\log(ab)$ has a rule, and why must you check every root of a logarithmic equation before you answer?
8. Your turn: the domain of $y = \log(5 - x)$, step 2
19. Your turn: a culture of $500$ cells grows continuously at $8\%$ an hour. Write its size after $t$ hours., step 2