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Great and complete logarithm functions properties cheat sheet
Typology: Cheat Sheet
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x f x = b , where b > 0 and x is any real
2 f x = x is NOT an exponential function.)
log (^) b x = y means that
y x = b where x > 0 , b > 0 , b ≠ 1
Think: Raise b to the power of y to obtain x. y is the exponent.
The key thing to remember about logarithms is that the logarithm is an exponent!
The rules of exponents apply to these and make simplifying logarithms easier.
Example: log 10 100 = 2 , since
2 100 = 10.
log 10 x is often written as just log x , and is called the COMMON logarithm.
log (^) e x is often written as ln x , and is called the NATURAL logarithm (note: e ≈ 2. 718281828459 ...).
Think: Multiply two numbers with the same base, add the exponents.
log (^) b = log b −log b log 8 1 7
log 8 56 log 87 log 8 = 8 =
Think: Divide two numbers with the same base, subtract the exponents.
P log (^) b = log log 100 3 log 100 3 2 6
3 = ⋅ = ⋅ =
Think: Raise an exponential expression to a power and multiply the exponents together.
b x
x log (^) b = log (^) b 1 = 0 (in exponential form, 1
0 b = ) ln 1 = 0
log (^) b b = 1 log 10 10 = 1 ln e = 1
b x
x log (^) b = x
x log 10 10 = e x
x ln =
b x b x =
log Notice that we could substitute y = log bx into the expression on the left
to form
y b. Simply re-write the equation y = log bx in exponential form
as
y x = b. Therefore, b b x b x y = =
log
. Ex: 26
ln 26 e =
b
a
a b log
log log = , for any positive base a. 0. 6476854
log 12
log 5 log 12 5 = ≈ ≈
This means you can use a regular scientific calculator to evaluate logs for any base.
Practice Problems contributed by Sarah Leyden, typed solutions by Scott Fallstrom
Solve for x (do not use a calculator).
2 9 x − =
2 1 3 =
x +
2 5 x −^ x + =
x
log (^) x 8 =−
1 2 2
2 1
2 2 x −^ x + =
2 (^33)
1 (^23)
(^1) x − x =
Solve for x , use your calculator (if needed) for an approximation of x in decimal form.
x
x x 5 = 9 ⋅ 4
x 10 =
− x e
x x 8 = 9
1 4 10 e
Solutions to the Practice Problems on Logarithms:
2 1 2 2 9 x −^ = ⇒ = x − ⇒ x = ⇒ x =±
2 1 15 2 1 3 =^ ⇒ = ⇒ + = ⇒ = ⇒ =
x x x
x x
3 x =^ ⇒ x = ⇒ x = 4.^ log^2525
2 5 x =^ ⇒ = x ⇒ x =
2 0 2 2 5 x −^ x + = ⇒ = x − x + ⇒ = x − x + ⇒ = x − x − ⇒ x = x =
3 3
3 3 =^ ⇒ 3 = ⇒ = ⇒ x = ⇒ x =
x x x
4
1 2
3 3 2 2 3 log 8 = − ⇒ = 8 ⇒ = 8 ⇒ =
− − x x x x
thenewequation. 3 istheonlysolution totheoriginal equation.
3 2 0 3 or 2 .Note: 2 isanextraneoussolution,whichsolvesonly
log log 1 1 log 1 6 6 0
2 2 2 6 6 6
x
x x x x x
x x x x x x x x
64
1 3 3 2 2 2 3 log 2 3 log 2 3 2 2
log log
(^1212) 2 1 (^12) = ⇒ = ⇒ = ⇒ = =
− − − x x x x
x
x
x
log log 3 8 1 log 1 2 6 16
2
2 (^223838)
2 2
2 2
x x x x x x
x x x x x
x x
x
729
(^61)
3 3 3
2 (^33)
1 (^23)
1
log log 1 log log 1 log 1 3
6 1
3 2 2 1
3 2
(^12) 3 2 2 1
− −
−
x x
x x
x x x x x
log 7
log 54 7 = 54 ⇒ x =log 7 54 ⇒ x = ≈
x
17 log 10 x = 17 ⇒ x = 10
4 4 5
5 4
5 = ⋅ ⇒ = ⇒ = ⇒ x = ⇒ x ≈
x x^ x x
x
x
− e x x
x
e x x e x e
8 8 9
9 = ⇒ = ⇒ x = ⇒ x =
x x^ x
10
1 4 4 4 4 4 = ⇒ + = ⇒ = − = − ⇒ = ≈
x + (^) e e x e x e e x
− − x x x