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 2 3 = 8,

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

Also, log 5 25 = 2, since 5 2 = 25

Log anything 1 = 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 log 10 A (pronounced "log base 10 of A").

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

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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 x 23 = 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 MN 3P rule #1 and rule #3. N 3 since we had 3 log a N, 3 is the exponent position. b) rules 2 & 3. Note the cube root and x 2 from the power 2/3. The minus sign indicates division.

We'll use these properties to solve log equations.

3/ Solve for x.

log 2 x + log 2 (x – 2) = 3

log 2 x (x – 2) = 3

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

2 3 = x(x – 2) = 8 since 2 3 = 8

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

We cannot take log 2 (–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.

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

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

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.CHANGE OF BASE:

Say we have to find log 3 12. 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.

log 3 12 = 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 b M where b is a base other than 10 or e, we divide log M by log b.

ex: in other words, 2 5 = 32

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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 =

2/

Evaluate

 a) log 2 (8 – 1 ) – log 3 9 + log ½ 4 b) 2 log 100 + 3 log 1000 – log 0.0001 c) log (1/ 3) 3 + 2 log x x 2 – 7 log ( 1/ n ) n 4

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3/ Write as a single log:

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

4/ Solve for x:

 a) log 2 (2x – 1) + log 2 (x – 1) = 0 b) log 4 (x – 3) – log 8 64 = log 4 (x + 1) c) log 3 x + log 3 (x + 6) = 3

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5/ Use Change of Base to find:

 a) log 7 19 b) log 2 9 c) log 8 (1 / 2) d) log 1 / 5 130

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

 a) log 45log (5 \$ 3 2 ) =log 5 + 2 log 3 = 1.6532 b) log 0.6log (3 / 5) = log 3 – log 5–0.2218 c) log 75log (5 2 \$ 3) =2 log 5 + log 3 = 1.8751 d) log 125 =log 5 3 = 3 log 52.0970

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2/

Evaluate

 a) log 2 (8 – 1 ) – log 3 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 x x 2 – 7 log ( 1/ n ) n 4 = –1 + 4 – 7(– 4) = 31

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3/ Write as a single log:

 a) 2 log x + 3 log y – 5 log z = log x 2 + log y 3 – log z 5 = b) a log b – 2 log c 2 + log d = log b a – log c 4 + log d =

4/ Solve for x:

 a) log 2 (2x – 1) + log 2 (x – 1) = 0 log 2 (2x – 1)(x – 1) = 0 log 2 (2x 2 – 3x + 1) = 0 2x 2 – 3x + 1 = 2 0 = 1 2x 2 – 3x = 0 becomes x (2x – 3) = 0, x = 0 or 1.5 if x = 0, 2x – 1 < 0 so x = 1.5 b) log 4 (x – 3) – log 8 64 = log 4 (x + 1) since log 8 64 = 2, we rewrite the equation: log 4 (x – 3) – log 4 (x + 1) = 2 so x – 3 = 16x + 16 or 15x = – 19 becomes x = – 19 / 15 no solution here. c) log 3 x + log 3 (x + 6) = 3 log 3 x (x + 6) = 3 log 3 (x 2 + 6x) = 3 becomes (x 2 + 6x) = 27 so x 2 + 6x – 27 = 0 becomes (x + 9)(x – 3) = 0 so x = 3 since –9 cannot be a solution

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5/ Use Change of Base to find:

 a) log 7 19 b) log 2 9 c) log 8 (1 / 2) d) log 1 / 5 130

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