Example 1: Solve the logarithmic equation
log_{2}(x  1) = 5.
Solution to example 1
 Rewrite the logarithm as an exponential using the definition.
x  1 = 2^{5}
 Solve the above equation for x.
x = 33
 check:
Left Side of equation
log_{2}(x  1) = log_{2}(33  1) = log_{2}(2^{5}) = 5
Right Side of equation = 5
 conclusion: The solution to the above equation is
x = 33
Example 2: Solve the logarithmic equation
log_{5}(x  2) + log_{5}(x + 2) = 1.
Solution to example 2
 Use the product rule to the expression in the right side.
log_{5}(x  2)(x + 2) = 1
 Rewrite the logarithm as an exponential (definition).
(x  2)(x + 2) = 5^{1}
 Which can be simplified as.
x^{2} = 9
 Solve for x.
x = 3 and x = 3
 check:
1st solution x = 3
Left Side of equation: log_{5}(3  2) + log_{5}(3 + 2) = log_{5}1 + log_{5}(3 + 2) = log_{5}5 = 1
Right Side of Equation
2nd solution x = 3
Left Side of equation: log_{5}(3  2) + log_{5}(3 + 2) = log_{5}(5) + log_{5}(1)
log_{5}(5) and log_{5}(1) are both undefined and therefore x = 3 is not a solution.
conclusion: The solutions to the given equation is x = 3
Example 3: Solve the logarithmic equation
log_{3}(x  2) + log_{3}(x  4) = log_{3}(2x^2 + 139)  1.
Solution to example 3
 We first replace 1 in the equation by log_{3}(3) and rewrite the equation as follows.
log_{3}(x  2) + log_{3}(x  4) = log_{3}(2x^2 + 139)  log_{3}(3)
 We now use the product and quotient rules of the logarithm to rewrite the equation as follows.
log_{3}[ (x  2)(x  4) ] = log_{3}[ (2x^2 + 139) / 3 ]
 Which gives the algebraic equation
(x  2)(x  4) = (2x^2 + 139) / 3
 Mutliply all terms by 3 and simplify
3(x  2)(x  4) = (2x^2 + 139)
 Solve the above quadratic equation to obtain
x = 5 and x = 23
 check:
1) x =  5 cannot be a solution to the given equation as it would make the argument of the logarithmic functions on the right negative.
2) x = 23
Right Side of equation:
log_{3}(23  2) + log_{3}(23  4) = log_{3}(21*19) = log_{3}(399)
Left Side of equation:
log_{3}(2(23)^2 + 139)  1 = log_{3}(1197)  log_{3}(3) = log_{3}(1197 / 3) = log_{3}(399)
 conclusion: The solution to the above equation is
x = 23
Example 4: Solve the logarithmic equation
log_{4}(x + 1) + log_{16}(x + 1) = log_{4}(8).
Solution to example 4
 We first note that 2 logarithms in the given equation have base 4 and one has base 16. We first use the change of base formula to write that
log_{16}(x + 1) = log_{4}(x + 1) / log_{4}(16) = log_{4}(x + 1) / 2 = log_{4}(x + 1)^{1/2}
 We now write the given equation as follows.
log_{4}(x + 1) + log_{4}(x + 1)^{1/2} = log_{4}(8)
 We use the product rule to write.
log_{4}(x + 1)(x + 1)^{1/2} = log_{4}(8)
 Which gives
(x + 1)(x + 1)^{1/2} = (8)
 which can be written as
(x + 1)^{3/2} = (8)
 Solve for x to obtain.
x = 3
 check:
Left Side of equation:
log_{4}(3 + 1) + log_{16}(3 + 1) = 1 + 1/2 = 3/2
Right Side of equation:
log_{4}(8) = log_{4}(4^{3/2}) = 3/2
 conclusion: The solution to the above equation is
x = 3
Example 5: Solve the logarithmic equation
log_{2}(x  4) + log_{sqrt(2)}(x^{3}  2) + log_{0.5}(x  4) = 20.
Solution to example 5
 We first use the change of base formula to write.
log_{sqrt(2)}(x^{3}  2) = log_{2}(x^{3}  2) / log_{2}(sqrt(2)) = 2log_{2}(x^{3}  2)
 We also use the change of base formula to write.
log_{0.5}(x  4) = log_{2}(x  4)
 Substitute into the equation and simplify the given equation.
2 log_{2}(x^{3}  2) = 20
 rewrite as.
log_{2}(x^{3}  2) = 10
 which gives
x^{3}  2 = 2^{10}
 Solve the above equation for x.
x = cube_root (1026)
Example 6: Solve the logarithmic equation
ln(x + 6) + log(x + 6) = 4.
Solution to example 6
 Use the change of base formula to rewrite log(x + 6) as
log(x + 6) = ln(x + 6) / ln(10)
 and substitute in the given equation
ln(x + 6) + ln(x + 6) / ln(10) = 4
 solve for ln(x + 6)
ln(x + 6) = 4 ln(10) / (1 + ln(10))
 solve the above for x
x = e^{4 ln(10) / (1 + ln(10))}  6
Example 7: Solve the logarithmic equation
log_{5}(ln(x + 3)  1) + log_{1/5}(ln(x + 3)  1) = 0.
Solution to example 7
 The change of base formula is used to write
log_{1/5}(ln(x + 3) = log_{5}(ln(x + 3)
 Substitute in the given equation
log_{5}(ln(x + 3)  1)  log_{5}(ln(x + 3)  1) = 0
 The left hand term is equal to 0 for x + 3 > 0 and ln(x + 3)  1 > 0.
x + 3 > 0 gives x > 3
 ln(x + 3)  1 > 0 gives.
ln(x + 3) > 1
 or
x + 3 > e
 or
x > e  3
 conclusion: The solution set to the above equation is given by the interval (e  3 , + infinity). It is an identity.
More topics to explore and Tests:
Experiment and Explore Mathematics: Tutorials and Problems
Solve Exponential and Logarithmic Equations
Solve Exponential and Logarithmic Equations  Tutorial
Logarithmic Functions (with applet)
