Logarithms and examples of working with logarithms were introduced on the logarithms page, and the exponential and logarithmic functions examples page.
Here on this how to simplify logarithms page we look further at situations of manipulating and solving where logarithms are present.
There are some rules of logarithms that can be used in certain examples where we want to express or solve for an unknown. These rules actually follow on from the exponent rules seen on the exponents in algebra page.
We can firstly recap the key rules for exponents.
Exponent Rules:
– a^m \times a^n \space {\small{=}} \space a^{m + n}
– {\Large{\frac{a^m}{a^n}}} \space {\small{=}} \space a^{m {\text{–}} n}
– (a^m)^n \space {\small{=}} \space a^{mn}
– a^0 \space {\small{=}} \space 1 , a^1 \space {\small{=}} \space a
The rules for logarithms follow in a similar pattern.
Logarithm Rules:
– {\tt{log}}_a (mn) \space {\small{=}} \space {\tt{log}}_a (m) + {\tt{log}}_a (n)
– {\tt{log}}_a (\frac{m}{n}) \space {\small{=}} \space {\tt{log}}_a (m) \space {\text{–}} \space {\tt{log}}_a (n)
– {\tt{log}}_a (x^n) \space {\small{=}} \space n \space {\tt{log}}_a (x)
– {\tt{log}}_a (a) \space {\small{=}} \space 1 , {\tt{log}}_a (1) \space {\small{=}} \space 0
Logarithm Rules Examples
(1.1)
a) Write {\tt{log}} (2a) as the sum of two logs.
Solution
{\tt{log}} (2a) \space {\small{=}} \space {\tt{log}}(2) + {\tt{log}}(a)
b) Write {\tt{log}}b^4 as a multiple.
Solution
{\tt{log}}b^4 \space {\small{=}} \space \space 4\space{\tt{log}}b
(1.2)
We will show all of the following written as the log of a single number.
a) {\tt{log}} (8) \space {\text{–}} \space {\tt{log}} (4) => {\tt{log}} (8) \space {\text{–}} \space {\tt{log}} (4) \space {\small{=}} \space {\tt{log}} (\frac{8}{4}) \space {\small{=}} \space {\tt{log}} (2)
b) 3{\tt{log}}4 => 3{\tt{log}}4 \space {\small{=}} \space {\tt{log}}4^3 \space {\small{=}} \space {\tt{log}}64
c) {\text{-}}2{\tt{log}}5 => {\text{-}}2{\tt{log}}5 \space {\small{=}} \space {\tt{log}}5^{{\text{-}}2} \space {\small{=}} \space {\tt{log}} {{\Large{\frac{1}{5^2}}}} \space {\small{=}} \space {\tt{log}} {{\Large{\frac{1}{25}}}}
d) 2{\tt{log}}2 \space {\text{–}} \space 3{\tt{log}}3 => 2{\tt{log}}2 \space {\text{–}} \space 3{\tt{log}}3 \space {\small{=}} \space {\tt{log}}2^2 \space {\text{–}} \space {\tt{log}}3^3 \space {\small{=}} \space {\tt{log}}4 \space {\text{–}} \space {\tt{log}}27 \space {\small{=}} \space {\tt{log}} \frac{4}{27}
(1.3)
Like in examples (1.2), we will show all of the following written as the log of a single number.
a) 1 + {\tt{log}}_3 (2) => 1 + {\tt{log}}_3 (2) \space {\small{=}} \space {\tt{log}}_3 (3) + {\tt{log}}_3 (2) \space {\small{=}} \space {\tt{log}}_3 (2 \times 3) \space {\small{=}} \space {\tt{log}}_3(6)
b) 2 + {\tt{log}}_{10} (5) => 2 + {\tt{log}}_{10} (5) \space {\small{=}} \space 2{\tt{log}}_{10} (10) + {\tt{log}}_{10} (5) \space {\small{=}} \space {\tt{log}}_{10}(10^2) + {\tt{log}}_{10}(5)
{\small{=}} \space {\tt{log}}_{10} (100) + {\tt{log}}_{10} (5) \space {\small{=}} \space {\tt{log}}_{10} (100 \times 5) \space {\small{=}} \space {\tt{log}}_{10}(500)
c) 1 + {\tt{log}}_5 b => 1 + {\tt{log}}_5 b \space {\small{=}} \space {\tt{log}}_5 5 + {\tt{log}}_5 b \space {\small{=}} \space {\tt{log}}_b (5 \times b) \space {\small{=}} \space {\tt{log}}_5 5b
(1.4)
Express {\tt{log}}_{10} (20) in terms of {\tt{log}}_{10} (2) and {\tt{log}}_{10} (5).
Solution
{\tt{log}}_{10} (20) \space {\small{=}} \space {\tt{log}}_{10} (4 \times 5) \space {\small{=}} \space {\tt{log}}_{10} (4) + {\tt{log}}_{10} (5)
{\small{=}} \space {\tt{log}}_{10} (2)^2 + {\tt{log}}_{10} (5) \space {\small{=}} \space 2{\tt{log}}_{10} (2) + {\tt{log}}_{10} (5)
(1.5)
Solve for x: {\tt{log}}x + 2{\tt{log}}2 \space {\small{=}} \space {\tt{log}}48.
Solution
{\tt{log}}x + 2{\tt{log}}2 \space {\small{=}} \space {\tt{log}}48 => {\tt{log}}x + {\tt{log}}2^2 \space {\small{=}} \space {\tt{log}}48 => {\tt{log}}x + {\tt{log}}4 \space {\small{=}} \space {\tt{log}}48
=> {\tt{log}}4x \space {\small{=}} \space {\tt{log}}48 , 4x \space {\small{=}} \space 48 \space \space , \space \space x \space {\small{=}} \space 12
How to Simplify Logarithms,
Change of Base Formula
When learning how to simplify logarithms, there is a formula that can be very helpful when we wish to establish the value of a given logarithm.
It is called the ‘change of base formula’.
Change of Base Formula => {\tt{log}}_a b \space {\small{=}} \space {\Large{\frac{{\tt{log}}_c b}{{\tt{log}}_c a}}}
The value of c can be any value we wish, and we can see in an example why this is helpful.
Example
(2.1)
Establish the value of {\tt{log}}_4 9 to 4 decimal places.
Solution
Now we know we can rewrite this log in the following way. {\tt{log}}_4 9 \space {\small{=}} \space {\Large{\frac{{\tt{log}}_c 9}{{\tt{log}}_c 4}}}
We just need to select a suitable value for c.
As the the common logarithm and the natural logarithm both have buttons on a standard calculator.
We can go with either {\tt{log}}_4 9 \space {\small{=}} \space {\Large{\frac{{\tt{log}}_{10} 9}{{\tt{log}}_{10} 4}}} or {\tt{log}}_4 9 \space {\small{=}} \space {\Large{\frac{{\tt{log}}_e 9}{{\tt{log}}_e 4}}}.
{\Large{\frac{{\tt{log}}_{10} 9}{{\tt{log}}_{10} 4}}} \space {\small{=}} \space 0.6309 , {\Large{\frac{{\tt{ln}} 9}{{\tt{ln}} 4}}} \space {\small{=}} \space 0.6309
We get the same result with the common and natural logarithm when using the change of base formula.