logs, rules, chofbase

Rules of Logs, Change of Base

Before the days of calculators, exponentiation, multiplication and division of huge numbers was a royal pain in the booty. However, the invention of logarithms changed exponentiation into multiplication, multiplication into addition, and division into subtraction.

Logs are exponents since, the log function is the inverse of the exponential function.

With the exponential function, the domain elements (x's) are exponents or powers,
the range (y's) elements are Real numbers.

With the log function, it's the opposite.

We input a number and extract the power of the base that gives us that number.

For example: log ₂ 8 = 3 since 2³ = 8,

or 3 is the exponent to which we must raise base 2 if we want to get 8.

Also, log₅ 25 = 2, since 5² = 25

Log_anything1 = 0 since (anything)⁰ = 1

Also, log_a a = 1

reminder: negative exponents turn things over, so negative logs do too.

example:

** Log Bases are never negative!! The domain of the log function is x > 0, so we can't take logs of negative numbers!! When solving log equations, we must check our solutions. Some of them won't work.

Log Bases

There are 2 standard bases for logs, the 2 you see on your calucuator.
They are 10 and the irrational number e, which, when you learn calculus, you'll understand.
For now just know that e l 2.71828... so
the graph of y = e^x will fall between y =2^x and y = 3^x .

Logs base e are called natural logs. ln A indicates log_e A -- pronounced "lawn A".

Logs base 10 are called common logs. log A indicates log₁₀ A (pronounced "log base 10 of A").

Notice, we only indicate the base with a subscripted value when it is neither e nor 10.

intro rules of logs change of base practice solutions

RULES OF LOGARITHMS

Since logs are exponents, they behave as such, ie:

RULE EXPLANATION

1) log _c (MN) = log _c M + log _c N.
so log _c 2y = log _c 2 + log _c y
log of a product = sum of the logs.
note: sum of logs not log of the sum!

2)
so
log of a quotient = difference of the logs.

3) log _c (M)ⁿ = n log _c M
so log _c x²³ = 23 log _c x
log of M to a power n = power n times log M

Examples

1/ Use the rules of logs to write these in expanded form.

a) log_a 6x²y³= log_a 6 + 2 log_a x + 3 log_a y use rule #1 and rule #3 since it's the log of a product and some terms include exponents.

b) we used all the rules here. Note the ½ from the
square root = power ½. 5 from log_c c = 1.

2/ Write as a single log.

a) log_a M + 3 log_a N + log_a P = log_a MN³P rule #1 and rule #3. N³ since we had 3 log_a N, 3 is the exponent position.

b) rules 2 & 3. Note the cube root and x² from the power 2/3. The minus sign indicates division.

We'll use these properties to solve log equations.

3/ Solve for x.

log₂ x + log ₂ (x – 2) = 3

log2 x (x – 2) = 3

Now we change to exponential form: (base to exponent power = number)

2³ = x(x – 2) = 8 since 2³ = 8

x² – 2x – 8 = 0 becomes (x – 4)(x + 2) = 0 so x = 4 or x = –2 .

We cannot take log₂ (–2) since there is no power of a positive number that will produce a negative value. The solution is x = 4.

To solve a log equation, we express the terms as a single log equal to a constant, then put it in exponential form and solve for the unknown.

4/ Given log 2 = 0.3010 and log 3 = 0.4771 find:

a) log 6
log (2 $ 3) =
log 2 + log 3 = 0. 7781
b) log 9
log 3² = 2 log 3
2(0.4771) = 0.9542
c) log 12
log (2² $ 3) =
2 log 2 + log 3 = 1.0791
d) log 2 / 3 =
log 2 – log 3 = – 0.1760

intro rules of logs change of base practice solutions

.CHANGE OF BASE:

Say we have to find log₃12. Our calculators only display logs in base 10 or base e. Let's solve the question as a log equation, then we'll have a change of base formula.

log₃12 = x in exponential form is 3^x = 12.

Taking log base 10 of both sides, we get x log 3 = log 12

Solving for x, we get .

This means that we must raise 3 to the power 2.26186 to get 12.
Try it! Enter 3 ^ 2.26186 and hit enter. If you don't get 12, you need a new calculator.

So, to find log_bM where b is a base other than 10 or e, we divide log M by log b.

ex: in other words, 2⁵ = 32

intro rules of logs change of base practice solutions

Practice

1/ Given log 3 = 0.4771 and log 5 = 0.6990 find:

a) log 45 b) log 0.6 c) log 75 d) log 125 =

Evaluate

a) log₂ (8^{– 1}) – log₃ 9 + log_½ 4 b) 2 log 100 + 3 log 1000 – log 0.0001

c) log_{(1/ 3)}3 + 2 log_xx² – 7 log_{( 1/}_n₎n⁴

3/ Write as a single log:

a) 2 log x + 3 log y – 5 log z b) a log b – 2 log c² + log d

4/ Solve for x:

a) log₂ (2x – 1) + log₂ (x – 1) = 0 b) log₄ (x – 3) – log₈ 64 = log₄ (x + 1)

c) log₃ x + log₃ (x + 6) = 3

5/ Use Change of Base to find:

a) log₇19 b) log₂9 c) log₈(1 / 2) d) log_{1 / 5}130

intro rules of logs change of base practice solutions

Solutions

1/ Given log 3 = 0.4771 and log 5 = 0.6990 find:

a) log 45
log (5 $ 3² ) =
log 5 + 2 log 3 = 1.6532
b) log 0.6
log (3 / 5) = log 3 – log 5
–0.2218
c) log 75
log (5² $ 3) =
2 log 5 + log 3 = 1.8751
d) log 125 =
log 5³ = 3 log 5
2.0970

Evaluate

a) log₂ (8^{– 1}) – log₃ 9 + log_½ 4
–3 – 2 – 2 = – 7
b) 2 log 100 + 3 log 1000 – log 0.0001
2(2) + 3(3) – (– 4) = 17

c) log_{(1/ 3)}3 + 2 log_xx² – 7 log_{( 1/}_n₎n⁴ =
–1 + 4 – 7(– 4) = 31

3/ Write as a single log:

a) 2 log x + 3 log y – 5 log z =
log x² + log y³ – log z⁵ =

b) a log b – 2 log c² + log d =
log b^a – log c⁴ + log d =

4/ Solve for x:

a) log₂ (2x – 1) + log₂ (x – 1) = 0
log₂ (2x – 1)(x – 1) = 0
log₂ (2x² – 3x + 1) = 0
2x² – 3x + 1 = 2⁰ = 1
2x² – 3x = 0 becomes x (2x – 3) = 0, x = 0 or 1.5
if x = 0, 2x – 1 < 0 so x = 1.5 b) log₄ (x – 3) – log₈ 64 = log₄ (x + 1)
since log₈ 64 = 2, we rewrite the equation:
log₄ (x – 3) – log₄ (x + 1) = 2

so x – 3 = 16x + 16 or 15x = – 19 becomes x = – 19 / 15
no solution here.

c) log₃ x + log₃ (x + 6) = 3
log₃ x (x + 6) = 3
log₃ (x² + 6x) = 3 becomes (x² + 6x) = 27
so x² + 6x – 27 = 0 becomes (x+ 9)(x – 3) = 0
so x = 3 since –9 cannot be a solution

5/ Use Change of Base to find:

a) log₇19
b) log₂9

c) log₈(1 / 2)
d) log_{1 / 5}130

intro rules of logs change of base practice solutions

Functions MathRom Index

RULE	EXPLANATION
1) log _c (MN) = log _c M + log _c N. so log _c 2y = log _c 2 + log _c y	log of a product = sum of the logs. note: sum of logs not log of the sum!
2) so	log of a quotient = difference of the logs.
3) log _c (M)ⁿ = n log _c M so log _c x²³ = 23 log _c x	log of M to a power n = power n times log M

a) log_a 6x²y³= log_a 6 + 2 log_a x + 3 log_a y	use rule #1 and rule #3 since it's the log of a product and some terms include exponents.
b)	we used all the rules here. Note the ½ from the square root = power ½. 5 from log_c c = 1.

a) log_a M + 3 log_a N + log_a P = log_a MN³P	rule #1 and rule #3. N³ since we had 3 log_a N, 3 is the exponent position.
b)	rules 2 & 3. Note the cube root and x² from the power 2/3. The minus sign indicates division.

a) log₂ (8^{– 1}) – log₃ 9 + log_½ 4	b) 2 log 100 + 3 log 1000 – log 0.0001
c) log_{(1/ 3)}3 + 2 log_xx² – 7 log_{( 1/}_n₎n⁴

a) log₂ (2x – 1) + log₂ (x – 1) = 0	b) log₄ (x – 3) – log₈ 64 = log₄ (x + 1)
c) log₃ x + log₃ (x + 6) = 3

a) log₂ (8^{– 1}) – log₃ 9 + log_½ 4 –3 – 2 – 2 = – 7	b) 2 log 100 + 3 log 1000 – log 0.0001 2(2) + 3(3) – (– 4) = 17
c) log_{(1/ 3)}3 + 2 log_xx² – 7 log_{( 1/}_n₎n⁴ = –1 + 4 – 7(– 4) = 31

a) log₂ (2x – 1) + log₂ (x – 1) = 0 log₂ (2x – 1)(x – 1) = 0 log₂ (2x² – 3x + 1) = 0 2x² – 3x + 1 = 2⁰ = 1 2x² – 3x = 0 becomes x (2x – 3) = 0, x = 0 or 1.5 if x = 0, 2x – 1 < 0 so *x = 1.5*	b) log₄ (x – 3) – log₈ 64 = log₄ (x + 1) since log₈ 64 = 2, we rewrite the equation: log₄ (x – 3) – log₄ (x + 1) = 2 so x – 3 = 16x + 16 or 15x = – 19 becomes x = – 19 / 15 no solution here.

c) log₃ x + log₃ (x + 6) = 3 log₃ x (x + 6) = 3 log₃ (x² + 6x) = 3 becomes (x² + 6x) = 27 so x² + 6x – 27 = 0 becomes (x+ 9)(x – 3) = 0 so x = 3 since –9 cannot be a solution