Logs

Section 1 of 7

$\log_{b} n = p \;\;\Longleftrightarrow\;\; b^{p} = n$

Worded problems

$P$ = Present Value

$F$ = Future Value

$t$ = time

$i = r$ = interest as a DECIMAL

APR = annual percentage rate = borrow

AER = annual equivalent rate = invest

Compound interest = per week = per month = per year

Section 2 of 7

Compound interest

Compounded annually

$F = P(1+i)^{t}$

$P = \dfrac{F}{(1+i)^{t}}$

Rule

$F = P(1+i)^{t}$

Compounded continuously

Number

$e = 2.71$

$F = Pe^{rt}$

$P = \dfrac{F}{e^{rt}}$ — continuous growth

Rule

$F = Pe^{rt}$

Worked example

€5,000 is invested for 3 years at 2% AER. Find value compounded:

(i) annually ? (ii) continuously ?

(i) Annually

$P = 5000, \;\; i = 0.02, \;\; t = 3$

$F = P(1+i)^{t}$

$F = 5000(1.02)^{3}$ = €5306.04

(ii) Continuously

$P = 5000, \;\; r = 0.02, \;\; t = 3$

$F = Pe^{rt}$

$F = 5000\, e^{0.02(3)} = 5000\, e^{0.06}$ = €5309.18

YOUR TURN

€2,000 invested for 4 years at 3% AER. Find value compounded
(i) annually ? (ii) continuously ?

Have a go on paper, then tap Show me for the working.

(i) $F = 2000(1.03)^{4}$ = €2251.02

(ii) $F = 2000\, e^{0.03(4)} = 2000\, e^{0.12}$ = €2254.99

(i) €2251.02 · (ii) €2254.99

Section 3 of 7

Finding the rate

Worked example

€5,000 amounts to €5,349 in 3 years at $r\%$ interest rate, compound continuously. Find $r$ to 1 decimal place.

$P = 5000, \;\; F = 5349, \;\; t = 3, \;\; r = ?$

$F = Pe^{rt}$

$5349 = 5000\, e^{3r}$

$\dfrac{5349}{5000} = e^{3r}$

$\log_{e}\dfrac{5349}{5000} = 3r$

$\ln\dfrac{5349}{5000} = 3r$

Notation

$\log_{e} x = \ln x$

$0.067 = 3r$

$r = 0.0224$

$r = 2.24\% \;\;\approx\;\; 2.2\%$

Depreciation

A tractor cost €256,000, sold off 5 years later at €160,000. Find rate of depreciation to 1 decimal place.

$P = 256{,}000, \;\; F = 160{,}000, \;\; t = 5, \;\; r = ?$

$F = Pe^{rt}$

$160{,}000 = 256{,}000\, e^{5r}$

$\dfrac{160}{256} = e^{5r}$

$\log_{e}\dfrac{160}{256} = 5r$

$\ln\dfrac{160}{256} = 5r$

$5r = -0.47$

$r = -0.094$

$r = 9.4\%$

Rule

$r > 0 \;\Rightarrow\;$ up
$r < 0 \;\Rightarrow\;$ down (decline / depreciation)

When is the tractor worth nothing?

$P = 256{,}000, \;\; r = -0.094, \;\; F = 0$

$0 = 256{,}000\, e^{-0.094\, t}$

$0 = e^{-0.094\, t}$

$\ln 0 = -0.094\, t$

Impossible. Answer: never.

Rule

$\log 0 = \ln 0 =$ impossible

YOUR TURN

€3,000 amounts to €3,600 in 4 years at $r\%$ continuously. Find $r$ to 1 decimal place.

Have a go on paper, then tap Show me for the working.

$3600 = 3000\, e^{4r}$
$\dfrac{3600}{3000} = e^{4r} \;\Rightarrow\; 1.2 = e^{4r}$
$\ln 1.2 = 4r$
$0.1823 = 4r \;\Rightarrow\; r = 0.0456$
$r = 4.6\%$

$r = 4.6\%$

Section 4 of 7

Bacteria growth

Worked example

Bacteria $B$ grows at a rate of 5% per day. Given $B = 200\, e^{rt}$. Find:

(i) Initial bacteria. (ii) Number of days for $B$ to reach 1000.

$r = 0.05, \;\; B = 200\, e^{rt}$

Rule

Initial $\;\Rightarrow\; t = 0$

(i) Initial bacteria

$B = 200\, e^{0} = 200$

(ii) Days to reach 1000

$B = 1000, \;\; r = 0.05$

$1000 = 200\, e^{0.05\, t}$

$5 = e^{0.05\, t}$

$\ln 5 = 0.05\, t$

$t = 32.18$

$t = 33$ days

Rule

Days are rounded up

YOUR TURN

Bacteria $B$ grows at 4% per day. $B = 300\, e^{rt}$. Find
(i) initial bacteria ? (ii) days for $B$ to reach 800.

Have a go on paper, then tap Show me for the working.

(i) $B = 300\, e^{0} = 300$

(ii) $800 = 300\, e^{0.04\, t}$
$\dfrac{8}{3} = e^{0.04\, t}$
$\ln \dfrac{8}{3} = 0.04\, t$
$0.9808 = 0.04\, t \;\Rightarrow\; t = 24.52$
$t = 25$ days

(i) $300$ · (ii) $25$ days

Section 5 of 7

Log equations: log = log

Log rules

$\log_{a}(xy) = \log_{a} x + \log_{a} y$
$\log_{a}\!\left(\dfrac{x}{y}\right) = \log_{a} x - \log_{a} y$
$\log_{a}(x^{q}) = q\,\log_{a} x$
$\log_{a} 1 = 0$

Notation

$\log x = \log_{10} x$
$\ln x = \log_{e} x$

$\log 2 + \log x = \log 8$

$\log 2x = \log 8$

$2x = 8 \;\Rightarrow\; x = 4$

$\log x - \log 3 = \log 4$

$\log \dfrac{x}{3} = \log 4$

$\dfrac{x}{3} = 4 \;\Rightarrow\; x = 12$

Goal

1 log = 1 log → drop the logs

Watch out: $\log\dfrac{x}{y} = \log x - \log y$, but $\dfrac{\log x}{\log y}$ ≠ $\log(x-y)$ — all three are different things.

$\log x + \log(x+2) = \log 8$

$\log x(x+2) = \log 8$

$x^{2} + 2x = 8$

$x^{2} + 2x - 8 = 0$

$(x+4)(x-2) = 0$

$x = -4 \;\;\text{(reject)}\;\;$ or $\;\; x = 2$

Rule

Cannot have $\log(\text{neg})$

$\log x + \log(x+5) = \log 14$

$\log x(x+5) = \log 14$

$x^{2} + 5x = 14$

$x^{2} + 5x - 14 = 0$

$(x-2)(x+7) = 0$

$x = 2 \;\;\text{or}\;\; x = -7 \;\;\text{(reject)}$

PATTERN CHECK

Solve $\log x + \log(x+3) = \log 10$

$\log x(x+3) = \log 10$
$x^{2} + 3x - 10 = 0$
$(x+5)(x-2) = 0$
$x = 2$ (reject $-5$)

$2\log x - \log(x+4) = \log 2$

$\log x^{2} - \log(x+4) = \log 2$

$\log \dfrac{x^{2}}{x+4} = \log 2$

$\dfrac{x^{2}}{x+4} = 2$

$x^{2} = 2x + 8$

$x^{2} - 2x - 8 = 0$

$x = 4 \;\;\text{or}\;\; x = -2 \;\;\text{(reject)}$

YOUR TURN

Solve $\log x + \log(x+6) = \log 16$

Have a go on paper, then tap Show me for the working.

$\log x(x+6) = \log 16$
$x^{2} + 6x - 16 = 0$
$(x+8)(x-2) = 0$
$x = 2$ (reject $-8$)

$x = 2$

YOUR TURN

Solve $2\log x - \log(x+3) = \log 4$

Have a go on paper, then tap Show me for the working.

$\log x^{2} - \log(x+3) = \log 4$
$\log \dfrac{x^{2}}{x+3} = \log 4$
$\dfrac{x^{2}}{x+3} = 4 \;\Rightarrow\; x^{2} = 4x + 12$
$x^{2} - 4x - 12 = 0$
$(x-6)(x+2) = 0$
$x = 6$ (reject $-2$)

$x = 6$

Section 6 of 7

Log equations: log = number

Master rule

$\log_{b} n = p \;\;\Longleftrightarrow\;\; b^{p} = n$

$\log_{2} x = 3$

$2^{3} = x \;\Rightarrow\; x = 8$

$\log(2x-1) = 2$

$\log_{10}(2x-1) = 2$

$10^{2} = 2x - 1$

$100 = 2x - 1$

$2x = 101 \;\Rightarrow\; x = 50.5$

PATTERN CHECK

Solve $\log_{3} x = 4$

$3^{4} = x$
$x = 81$

$\log_{2}(3x-1) = 3$

$2^{3} = 3x - 1$

$8 = 3x - 1 \;\Rightarrow\; x = 3$

$\log_{3} \dfrac{x}{x-1} = 2$

$3^{2} = \dfrac{x}{x-1}$

$9 = \dfrac{x}{x-1}$

$9x - 9 = x$

$8x = 9 \;\Rightarrow\; x = \dfrac{9}{8}$

$\log_{3}(7x-1) = -1$

$3^{-1} = 7x - 1$

$\dfrac{1}{3} = 7x - 1$

$x = \dfrac{4}{21}$

YOUR TURN

Solve $\log_{5}(2x+3) = 2$

Have a go on paper, then tap Show me for the working.

$5^{2} = 2x + 3$
$25 = 2x + 3$
$2x = 22 \;\Rightarrow\; x = 11$

$x = 11$

$\log_{3}(x-2) = 2$

$3^{2} = x - 2$

$11 = x$

Combine, then use the master rule

$\log_{2} x + \log_{2}(x-2) = 3$

$\log_{2} x(x-2) = 3$

$x^{2} - 2x = 2^{3}$

$x^{2} - 2x - 8 = 0$

$(x-4)(x+2) = 0$

$x = 4 \;\;\text{or}\;\; x = -2 \;\;\text{(reject)}$

$2\log_{2} x - \log_{2}(x-1) = 2$

$\log_{2} x^{2} - \log_{2}(x-1) = 2$

$\log_{2} \dfrac{x^{2}}{x-1} = 2$

$2^{2} = \dfrac{x^{2}}{x-1}$

$4(x-1) = x^{2}$

$x^{2} - 4x + 4 = 0$

$(x-2)(x-2) = 0$

$x = 2$

YOUR TURN

Solve $\log_{3} x + \log_{3}(x-6) = 3$

Have a go on paper, then tap Show me for the working.

$\log_{3} x(x-6) = 3$
$x^{2} - 6x = 3^{3} = 27$
$x^{2} - 6x - 27 = 0$
$(x-9)(x+3) = 0$
$x = 9$ (reject $-3$)

$x = 9$

Section 7 of 7

Change of base

$\log_{b} a = \dfrac{\log_{c} a}{\log_{c} b}$

Change $\log_{8} x$ to base 2.

$\log_{8} x = \dfrac{\log_{2} x}{\log_{2} 8} = \dfrac{\log_{2} x}{3}$

Change $\log_{25} x$ to base 5.

$\log_{25} x = \dfrac{\log_{5} x}{\log_{5} 25} = \dfrac{\log_{5} x}{2}$

$= \dfrac{1}{2} \log_{5} x = \log_{5} x^{1/2} = \log_{5} \sqrt{x}$

Change $\log_{x} 27$ to base 3.

$\log_{x} 27 = \dfrac{\log_{3} 27}{\log_{3} x} = \dfrac{3}{\log_{3} x}$

PATTERN CHECK

Change $\log_{4} x$ to base 2.

$\log_{4} x = \dfrac{\log_{2} x}{\log_{2} 4} = \dfrac{\log_{2} x}{2}$

YOUR TURN

Change $\log_{27} x$ to base 3.

Have a go on paper, then tap Show me for the working.

$\log_{27} x = \dfrac{\log_{3} x}{\log_{3} 27} = \dfrac{\log_{3} x}{3}$

$\dfrac{\log_{3} x}{3}$

Change $\log_{5} x$ to base $x$.

$\log_{5} x = \dfrac{\log_{x} x}{\log_{x} 5} = \dfrac{1}{\log_{x} 5}$

Rule

$\log_{a} a = 1$

Mixed bases

$\log_{4}(3x+1) = \log_{2}(x-1)$

$\log_{4}(3x+1) = \dfrac{\log_{2}(3x+1)}{\log_{2} 4}$

$\dfrac{\log_{2}(3x+1)}{2} = \log_{2}(x-1)$

$\log_{2}(3x+1) = 2\log_{2}(x-1)$

$\log_{2}(3x+1) = \log_{2}(x-1)^{2}$

$3x + 1 = x^{2} - 2x + 1$

$x^{2} - 5x = 0$

$x(x-5) = 0$

$x = 0 \;\;\text{(reject)}\;\;$ or $\;\; x = 5$

YOUR TURN

Solve $\log_{9}(2x+1) = \log_{3}(x-1)$

Have a go on paper, then tap Show me for the working.

$\log_{9}(2x+1) = \dfrac{\log_{3}(2x+1)}{2}$
$\dfrac{\log_{3}(2x+1)}{2} = \log_{3}(x-1)$
$\log_{3}(2x+1) = \log_{3}(x-1)^{2}$
$2x + 1 = x^{2} - 2x + 1$
$x^{2} - 4x = 0 \;\Rightarrow\; x(x-4) = 0$
$x = 4$ (reject $0$)

$x = 4$

Substitution

$\log_{2} x + 2\log_{x} 2 = 3$

$\log_{x} 2 = \dfrac{\log_{2} 2}{\log_{2} x} = \dfrac{1}{\log_{2} x}$

$\log_{2} x + \dfrac{2}{\log_{2} x} = 3$

$(\log_{2} x)^{2} + 2 = 3 \log_{2} x$

Rule

$\log a^{n} = n \log a$
$(\log a)^{n} \;\ne\; n \log a$

Let $t = \log_{2} x$

$t^{2} + 2 = 3t$

$t^{2} - 3t + 2 = 0$

$(t-1)(t-2) = 0$

$t = 1 \;\;\text{or}\;\; t = 2$

$\log_{2} x = 1 \;\Rightarrow\; x = 2^{1} = 2$

$\log_{2} x = 2 \;\Rightarrow\; x = 2^{2} = 4$

YOUR TURN

Solve $\log_{3} x + 6\log_{x} 3 = 5$

Have a go on paper, then tap Show me for the working.

$\log_{x} 3 = \dfrac{1}{\log_{3} x}$
$\log_{3} x + \dfrac{6}{\log_{3} x} = 5$
Let $t = \log_{3} x$:
$t^{2} + 6 = 5t$
$t^{2} - 5t + 6 = 0$
$(t-2)(t-3) = 0 \;\Rightarrow\; t = 2$ or $t = 3$
$x = 3^{2} = 9$ or $x = 3^{3} = 27$

$x = 9$ or $x = 27$

That’s Logs.

Every rule built from your handwritten PDF, with a 4-beat rhythm: derive, ask, tell, do.