Module 3-­‐ Exponents and Logarithms notes and examples For more problems and explanations, view these excellent videos: https://www.khanacademy.org/math/algebra2/logarithms-­‐tutorial Properties of Exponents: 1. ( x a ) ( x b ) = x a+b
2.
xa
= x a−b
xb
3. ( x a ) = x ab
b
Properties of Logarithms: note-­‐ in each of the following rules, they are valid ONLY for a,b,c, x > 0 . When used as a base of the logarithm, the base cannot equal 1 either. So it is important to check your solutions when working with logarithms. 1. log b a + log b c = log b ac
⎛ a⎞
2. log b a − log b c = log b ⎜ ⎟
⎝ c⎠
3. log b a n = n log b a
log c a
4. log b a =
(change of base)
log c b
5. b logb x = x
6. Special base: ln x = ln e x where e  2.718 so by using 'ln' the base is assumed to be e.
Sample Problems 1. Solve: 2 − x+1 = 4(8 x )
2 − x+1 = 2 2 (2 3x )
2 − x+1 = 2 2+3x
− x + 1 = 2 + 3x (bases equal so exponents must be equal)
− 4x = 1
1
x=−
4
2. Solve for x :
9 x − 3x+1 = −2
32 x − 3x × 31 + 2 = 0
Let y = 3x
y 2 − 3y + 2 = 0
(y − 2)(y − 1) = 0
y = 2,1
3x = 2 or 3x = 1
x ln 3 = ln 2 or x = 0
ln 2
x=
or x = 0
ln 3
ln 2
note:
= log 3 2 so you need to know both forms of the same answer,
ln 3
especially in multiple choice. Other correct form of the answer is also
1
log 2 3
3. log(9x + 1) − log(x − 1) = 1
⎛ 9x + 1 ⎞
log ⎜
=1
⎝ x − 1 ⎟⎠
9x + 1
= 101
x −1
9x + 1 = 10(x − 1)
9x + 1 = 10x − 10
x = 11 and the answer checks 4. Solve: log x (x + 1) = 2
Solution: x 2 = x + 1
Using the quadratic formula, x =
x=
1 ± 1+ 4
2
since x > 0 reject x =
∴x =
−b ± b 2 − 4ac
2a
1+ 5
only
2
1− 5
(remember to check all answers for validity)
2
5. Solve for the point of intersection of h(x) = 1+ log 4 x and p(x) = log16 x
Solution:
Let h(x) = p(x)
1+ log 4 x = log16 x
log 4 x
1+ log 4 x =
log 4 16
log 4 x
1+ log 4 x =
2
2 + 2 log 4 x = log 4 x
log 4 x = −2
x = 4 −2
x=
1
16
1
1
, substitute into either function y = log16
= −1
16
16
⎛ 1
⎞
The point of intersection is ⎜ ,−1⎟
⎝ 16 ⎠
6. If log 2 3 = a and log 2 5 = b, find log10 0.12 in terms of a and b
If x =
Solution:
⎛ 12 ⎞
log10 0.12 = log10 ⎜
= log10 12 − log10 100 = log10 12 − 2
⎝ 100 ⎟⎠
As the original problem refers to base 2, convert log10 12 to base 2 using change of
base formula.
log 2 12
log 2 3 + log 2 4
a+2
=
−2 =
−2 =
−2
log 2 10
log 2 5 + log 2 2
b +1
We will usually need to write as one fraction, so
=
a+2
a+2
⎛ b +1⎞ a − 2b
−2 =
− 2⎜
=
⎝ b +1⎟⎠
b +1
b +1
b +1
7. Solve e x = 5 x−1. Leave answers exact.
x = (x − 1)ln 5
x = x ln 5 − ln 5
x(ln 5 − 1) = ln 5
ln 5
− ln 5
x=
or
(exact)
ln 5 − 1 1− ln 5