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Solving Exponential and Logarithmic Equations

Solving Exponential and Logarithmic Equations

Solving Exponential and Logarithmic Equations

Solving Exponential and Logarithmic Equations

Solving Exponential and Logarithmic Equations

Solving Exponential and Logarithmic Equations

Solving Equations Containing Variable Exponents

To solve an equation containing a variable exponent, isolate the exponential quantity. Then take a logarithm, to the base of the exponent, of both sides.


Example 1: Solve for x : 3x = 15 .

3x = 15
log33x = log315
x = log315
x =
x 2.465


Example 2: Solve for x : 4·52x = 64 .

4·52x = 64
52x = 16
log552x = log516
2x = log516
2x =
2x 1.723
x 0.861

Solving Equations Containing Logarithms

To solve an equation containing a logarithm, use the properties of logarithms to combine the logarithmic expressions into one expression. Then convert to exponential form and evaluate. Check the solution(s) and eliminate any extraneous solutions--recall that we cannot take the logarithm of a negative number.

Example 1: Solve for x : log3(3x) + log3(x - 2) = 2 .

log3(3x) + log3(x - 2) = 2
log3(3x(x - 2)) = 2
32 = 3x(x - 2)
9 = 3x 2 - 6x
3x 2 - 6x - 9 = 0
3(x 2 - 2x - 3) = 0
3(x - 3)(x + 1) = 0
x = 3, - 1
Check:

  • x = 3 : log3(3·3) + log31 = 2 + 0 = 2 . x = 3 is a solution.
  • x = - 1 : log3(3· -1) + log3(- 1 - 2) = log3(- 3) + log3(- 3) does not exist. x = - 1 is not a solution.
Thus, x = 3 .


Example 2: Solve for x : 2 log(2x+1)(2x + 4) - log(2x+1)4 = 2 .

2 log(2x+1)(2x + 4) - log(2x+1)4 = 2
log(2x+1)(2x + 4)2 - log(2x+1)4 = 2
log(2x+1) = 2
(2x + 1)2 =
(2x + 1)2 =
4x 2 +4x + 1 = x 2 + 4x + 4
3x 2 - 3 = 0
3(x 2 - 1) = 0
3(x + 1)(x - 1) = 1
x = 1, - 1
Check:

  • x = 1 : 2 log36 - log34 = log362 - log34 = log3 = log39 = 2 . x = 1 is a solution.
  • x = - 1 : 2 log-12 - log-14 does not exist (the base cannot be negative).
Thus, x = 1 .