# Solve Logarithmic Equations - Detailed Solutions

Solve logarithmic equations including some challenging questions. Detailed solutions are presented. The logarithmic equations in examples 4, 5, 6 and 7 involve logarithms with different bases and are therefore challenging.

 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)