Properties of Logarithms and Solving Logarithmic Equations

In this section we will be working with Properties of Logarithms in an
attempt to take equations with more than one logarithm and condense
them down into just one single logarithm.
Properties of Logarithms:
a. Product Rule:
log 𝑐 (π‘Ž) + log 𝑐 (𝑏) = log 𝑐 (π‘Ž βˆ™ 𝑏)
When two or more logarithms with the same base are added, those
logarithms can be condensed into one logarithm whose argument is
the product of the original arguments (π‘Ž βˆ™ 𝑏 in the example above)
Order is not important when multiplying or adding, so changing the order
of the factors in an argument or changing the order of the terms being
added together does not change the answer.
b. Quotient Rule:
π‘Ž
log 𝑐 (π‘Ž) βˆ’ log 𝑐 (𝑏) = log 𝑐 ( )
𝑏
When two or more logarithms with the same base are subtracted,
those logarithms can be condensed into one logarithm whose
argument is the quotient of the original arguments
π‘Ž
( 𝑖𝑛 π‘‘β„Žπ‘’ 𝑒π‘₯π‘Žπ‘šπ‘π‘™π‘’ π‘Žπ‘π‘œπ‘£π‘’)
𝑏
Order is important when dividing and subtracting; the numerator is always
the argument of the first log term listed (or the argument of the term listed
first is always the numerator).
c. Power Rule:
𝑝 βˆ™ log 𝑐 (π‘Ž) = log 𝑐 (π‘Žπ‘ )
A factor times a logarithm can be re-written as the argument of the
logarithm raised to the power of that factor
There is no Property of Logarithms that applies when the argument is a
sum or difference.
log c (π‘Ž + 𝑏) β‰  log c (π‘Ž) + log c (𝑏)
log c (π‘Ž βˆ’ 𝑏) β‰  log c (π‘Ž) βˆ’ log c (𝑏)
If an equation has more than one logarithm on either side of the equation,
use the Properties of Logarithms to simplify as much as possible, then
solve by converting to exponential form. REMEMBER THAT YOU
MUST CHECK YOUR ANSWERS WHEN SOLVING LOG
EQUATIONS TO VERIFY THAT THEY MAKE THE ORIGINAL
ARGUMENTS POSITIVE.
Example 1: Solve each of the following logarithmic equations and
CHECK YOUR SOLUTIONS.
a. 2 = log 3 (π‘₯) βˆ’ log 3 (π‘₯ βˆ’ 1)
b. log 2 (π‘₯ + 7) + log 2 (π‘₯ ) = 3
c. 2 ln(π‘₯ ) βˆ’ ln(5π‘₯ βˆ’ 6) = 0
d. log(1 βˆ’ π‘₯ ) = 1 βˆ’ log(π‘₯ βˆ’ 1)
1
e. 2 log(π‘₯ + 1) = log(1 βˆ’ π‘₯ )
Example 2: Solve each of the following logarithmic equations and
CHECK YOUR SOLUTIONS.
a. log 5 (10) βˆ’ log 5 (50) = log 5 (0.5) + log 5 (4π‘₯ + 6)
b. log 4 (3π‘₯ + 2) = log 4 (5) + log 4 (3)
c. ln(π‘₯ βˆ’ 2) = 1 βˆ’ ln(π‘₯)
(hint: when you get a quadratic equation, solve by completing the
square rather than using the quadratic formula)
d. log(π‘₯ + 2) βˆ’ log(π‘₯) = 2 log(4)
e. log 3 (π‘₯ + 3) + log 3 (π‘₯ + 5) = 1
f. log(4 βˆ’ 5π‘₯ ) βˆ’ log(10 βˆ’ π‘₯ 2 ) = 0
g. ln(π‘₯ ) + ln(6π‘₯ βˆ’ 5) = 0
h. 2 log 8 (π‘₯) = 3 log 8 (4)
Remember that there is no Property of Logarithms that applies when the
argument is a sum or difference.
log c (π‘Ž + 𝑏) β‰  log c (π‘Ž) + log c (𝑏)
log c (π‘Ž βˆ’ 𝑏) β‰  log c (π‘Ž) βˆ’ log c (𝑏)
5
Therefore equations such as ln(π‘₯ 2 + 1) = 2 or log (π‘₯2 βˆ’ 4) = 0
CANNOT be simplified using any Properties of Logarithms first before
solving. Instead, these types of equations would simply be solved by
converting to exponential form.
Example 3: Solve each of the following logarithmic equations and
CHECK YOUR SOLUTIONS.
a. ln(π‘₯ 2 + 1) = 2
b. ln(π‘₯ 2 + 1) = βˆ’1
5
c. log (π‘₯ 2 βˆ’ 4) = 0
Answers to Examples:
9
1a. π‘₯ = 8 ; 1b. π‘₯ = 1 ; 1c. π‘₯ = 2; π‘₯ = 3 ; 1d. 𝑁𝑂 π‘†π‘‚πΏπ‘ˆπ‘‡πΌπ‘‚π‘ ;
1e. π‘₯ = 0 ;
7
13
2
2a. π‘₯ = βˆ’ 5 ; 2b. π‘₯ = 3 ; 2c. π‘₯ = 1 + βˆšπ‘’ + 1 ; 2d. π‘₯ = 15 ;
2e. π‘₯ = βˆ’2 ; 2f. π‘₯ = βˆ’1 ; 2g. π‘₯ = 1 ; 2h. π‘₯ = 8 ;
3
3a. π‘₯ = ±βˆšπ‘’2 βˆ’ 1 ; 3b. 𝑁𝑂 π‘†π‘‚πΏπ‘ˆπ‘‡πΌπ‘‚π‘ ; 3c. π‘₯ = ± 2;