Example 1: Solve the logarithmic equation
log2(x - 1) = 5.
Solution to example 1
- Rewrite the logarithm as an exponential using the definition.
x - 1 = 25
- Solve the above equation for x.
x = 33
- check:
Left Side of equation
log2(x - 1) = log2(33 - 1) = log2(25) = 5
Right Side of equation = 5
- conclusion: The solution to the above equation is
x = 33
Example 2: Solve the logarithmic equation
log5(x - 2) + log5(x + 2) = 1.
Solution to example 2
- Use the product rule to the expression in the right side.
log5(x - 2)(x + 2) = 1
- Rewrite the logarithm as an exponential (definition).
(x - 2)(x + 2) = 51
- Which can be simplified as.
x2 = 9
- Solve for x.
x = 3 and x = -3
- check:
1st solution x = 3
Left Side of equation: log5(3 - 2) + log5(3 + 2) = log51 + log5(3 + 2) = log55 = 1
Right Side of Equation
2nd solution x = -3
Left Side of equation: log5(-3 - 2) + log5(-3 + 2) = log5(-5) + log5(-1)
log5(-5) and log5(-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
log3(x - 2) + log3(x - 4) = log3(2x^2 + 139) - 1.
Solution to example 3
- We first replace 1 in the equation by log3(3) and rewrite the equation as follows.
log3(x - 2) + log3(x - 4) = log3(2x^2 + 139) - log3(3)
- We now use the product and quotient rules of the logarithm to rewrite the equation as follows.
log3[ (x - 2)(x - 4) ] = log3[ (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:
log3(23 - 2) + log3(23 - 4) = log3(21*19) = log3(399)
Left Side of equation:
log3(2(23)^2 + 139) - 1 = log3(1197) - log3(3) = log3(1197 / 3) = log3(399)
- conclusion: The solution to the above equation is
x = 23
Example 4: Solve the logarithmic equation
log4(x + 1) + log16(x + 1) = log4(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
log16(x + 1) = log4(x + 1) / log4(16) = log4(x + 1) / 2 = log4(x + 1)1/2
- We now write the given equation as follows.
log4(x + 1) + log4(x + 1)1/2 = log4(8)
- We use the product rule to write.
log4(x + 1)(x + 1)1/2 = log4(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:
log4(3 + 1) + log16(3 + 1) = 1 + 1/2 = 3/2
Right Side of equation:
log4(8) = log4(43/2) = 3/2
- conclusion: The solution to the above equation is
x = 3
Example 5: Solve the logarithmic equation
log2(x - 4) + logsqrt(2)(x3 - 2) + log0.5(x - 4) = 20.
Solution to example 5
- We first use the change of base formula to write.
logsqrt(2)(x3 - 2) = log2(x3 - 2) / log2(sqrt(2)) = 2log2(x3 - 2)
- We also use the change of base formula to write.
log0.5(x - 4) = -log2(x - 4)
- Substitute into the equation and simplify the given equation.
2 log2(x3 - 2) = 20
- rewrite as.
log2(x3 - 2) = 10
- which gives
x3 - 2 = 210
- 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 = e4 ln(10) / (1 + ln(10)) - 6
Example 7: Solve the logarithmic equation
log5(ln(x + 3) - 1) + log1/5(ln(x + 3) - 1) = 0.
Solution to example 7
- The change of base formula is used to write
log1/5(ln(x + 3) = -log5(ln(x + 3)
- Substitute in the given equation
log5(ln(x + 3) - 1) - log5(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)
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