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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intro  rules of logs  change of base  practice  solutions 
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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^{ 3}P  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 6 log (2 $ 3) = log 2 + log 3 = 0. 7781 
b) log 9 log 3^{ 2} = 2 log 3 2(0.4771) = 0.9542 
c) log 12 log (2^{ 2} $ 3) = 2 log 2 + log 3 = 1.0791 
d) log 2 / 3 = log 2 – log 3 = – 0.1760 
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intro  rules of logs  change of base  practice  solutions 
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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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intro  rules of logs  change of base  practice  solutions 
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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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intro  rules of logs  change of base  practice  solutions 
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1/ Given log 3 = 0.4771 and log 5 = 0.6990 find:
a) log 45 log (5 $ 3^{ 2} ) = log 5 + 2 log 3 = 1.6532 
b) log 0.6 log (3 / 5) = log 3 – log 5 –0.2218 
c) log 75 log (5^{ 2} $ 3) = 2 log 5 + log 3 = 1.8751 
d) log 125 = log 5^{ 3} = 3 log 5 2.0970 
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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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intro  rules of logs  change of base  practice  solutions 
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