Here are solutions to the practice problems: log(2x – 4) = 3 2x – 4 = 23 = 8 2x – 4 = 8 2x = 12 x = 6 log2(x) + log2 (x – 6) = 4 log2(x(x – 6)) = 4 x(x – 6) = 24 x2 – 6x – 24 = 0 (x – 6)(x + 4) = 0 x = 6, x = -4 log4(x + 4) – log4(x – 1) = 2 log4((x + 4)/(x – 1)) = 2 (x

2. Learning Intention
Recap logarithmic expressions, rules & equations
Success criteria: you will be able to
Express log statements in exponential form
Apply log rules
Solve log equations

3. Back to Basics
A log is just the inverse of an exponential!
y = bx is equivalent to logb(y) = x
(means the exact same thing as)
Image source: www.purplemath.com

4. The Log Switcheroo
Write the following exponential expressions in log form:
63 = 216
log6(216) = 3
45 = 1024 log4(1024) = 5
Write the following logarithmic expressions in
exponential form:
log2(8) = 3 23 = 8
52 = 25
log5(25) = 2 641/3 = 4
log64(4) = 1/3

5. SIMPLE (base 10) EXAMPLES
Exponential
Number Logarithm
Expression
1000 103 3
100 102 2
10 101 1
1 100 0
1/10 = 0.1 10-1 -1
1/100 = 0.01 10-2 -2
1/1000 = 0.001 10-3 -3

6. Some Things To
Remember
b1 = b , so logb(b) = 1, for any base b
b0 = 1 , so logb(1) = 0
logb(a) is undefined if a is negative
logb(0) is undefined
logb(bn) = n

7. Calculations with Logs
Because logarithms are exponents, mathematical operations
involving them follow the same rules as those for exponents:
1) Multiplication inside the log can be turned into addition outside
the log, and vice versa.
logb(mn) = logb(m) + logb(n)
2) Division inside the log can be turned into subtraction outside the
log, and vice versa.
logb(m/n) = logb(m) – logb(n)
3) An exponent on everything inside a log can be moved out front as a
multiplier, and vice versa.
logb(mn) = n · logb(m)

8. Log Rule Practice
Expand log3(2x)
log3(2x) = log3(2) + log3(x)
Expand log4( 16/x )
log4( 16/x ) = log4(16) – log4(x)
log4(16) = 2 so log4( 16/x ) = 2 – log4(x)
Expand log5(x3)
log5(x3) = 3 · log5(x) = 3log5(x)

9. Change of Base
eg Evaluate log3(6)
log3(6) = log(6) = 1.63092975...
log(3)

10. Solving a Log Equation
Step 1: Write as one log on one side
Step 2: Use the definition of logarithms to write in
exponential form (or vice versa)
Step 3: Solve for x
eg log5(x+2) = 3
53 = x + 2
125 = x + 2
x = 123

11. A Trickier Example
Write as one log on one log(x + 21) + log(x) = 2
side log [(x+21)x] = 2 that when there is no base written on a lo
Remember
Use the definition of x2+21x = 102
logarithms to write in
exponential form x2+21x – 100 = 0
(x+25)(x-4) = 0
Solve for x
x = 4, -25
BUT we CANNOT take the log of a
negative number, so we will have
to throw out x = -25 as one of our
solutions

12. Exam Question
Write as one log on one H(t) = 3 + (1.24)t
side
When does H = 7?
Use the definition of
7 = 3 + (1.24)t
logarithms to write in
exponential form (or 4 = (1.24)t
vice versa)
log (4) = log (1.24t)
Solve for x
log (4) = t x log (1.24)
t= log (4) = 6.44 years
log(1.24)

13. Try the Following:
Write as one log on one log(2x – 4) = 3
side
log2(x) + log2 (x – 6) = 4
Use the definition of
logarithms to write in log4(x + 4) – log4(x – 1) = 2
exponential form
Solve for x