Solve Exponential and Logarithmic Equations - Tutorial
Tutorials on how to solve exponential and logarithmic equations with examples and detailed solutions are presented. A tutorials with exercises and solutions on the use of the rules of logarithms and exponentials may be useful before you start the present tutorial.
Review of the Properties and Rules of Logarithm and Exponentials
Some of the most important rules of logarithms and exponentials and properties are listed below.- logb(U V) = logb U + logb V
- logb(U / V) = logb U - logb V
- logb (Ur) = r logb U
- logb (bx) = x
- blogb x = x , x > 0
- logb U = logb V ⇔ U = V
- a x a y = a x + y
-
a x / a y = a x - y
Being inverse of each other, logarithmic and exponential functions to the same base are related as follows: -
logy x = y ⇔ x = b y
This last inverse function property helps in converting exponential equation to a logarithmic one and a logarithmic equation to an exponential one.
Examples with Solutions
Example 1
ln (x) = 5 and check the solution found.Solution to Example 1 Use the inverse property (9) given above to rewrite the given logarithmic (ln has base e) equation as follows: x = e 5 Check Solution Substitute x by e 5 in the left side of the given equation and simplify ln (e 5) = 5 , use property (4) to simplify which is equal to the right side. Hence x = e 5 is the solution to the given equation.
Example 2
e x = 6 and check the solution found.Solution to Example 2 Use the inverse property (9) given above to rewrite the given exponential equation as follows: x = ln(6) Check Solution Substitute x by ln(6) in the left hand side of the equation e x and simplify e ln(6) = 6 , use property (5) to simplify which is equal to the right hand side of the given equation. Hence x = ln(6) is the solution to the given equation.
Example 3
ln (x) + ln (2) = 3 Solution to Example 3 Use property (1) from right to left to group the two ln terms on the left ln(2 x) = 3 Use property (9) to rewrite the above logarithmic equation as follows: 2 x = e 3 Solve for x x = e 3 / 2
Example 4
e 3x = - 1 Solution to Example 4 The range of basic exponential functions is (0 , + ∞), hence e 3x cannot be negative and therefore the given equation has no real solutions.
Example 5
Solve the logarithmic equation and check the solution obtained.
ln(x) + 2 = - 3 ln (x) + 10 Solution to example 5 combine like terms 4 ln(x) = 8 Divide both sides by 4. ln(x) = 2 Use the inverse property (9) to rewrite the above using exponentials. x = e 2 check solution obtained Evaluate left side of the given equation: ln(e 2) + 2 = 2 + 2 = 4 Evaluate right side of equation: - 3 ln(e 2) + 10 = -3(2) + 10 = 4 Hence, the solution to the above equation is x = e 2.
Example 6
2 e x + e - x = 3 Solution to Example 6multiply all terms by e x (2 e x + e - x)e x = 3e x Expand 2 e x e x + e - x e x = 3 e x Use exponents properties to simplify. 2 e 2 x + 1 = 3 e x note that e 2 x = (e x) 2 The equation may be written as 2 (e x) 2 + 1 = 3 e x Use substitution: let u = e x and rewrite the equation in u 2 u 2 + 1 = 3u rewrite the equation in standard form 2 u 2 - 3 u + 1 = 0 solve, for u, the above quadratic equation u = 1 , u = 1/2 We substitute u by e x in the above solutions e x = 1 and e x = 1/2 Solve, for x, the first of the above equations e x = 1 Use inverse property (9) to rewrite the above in logarithm form (or take logarithms of both sides) x = ln(1) = 0 Solve, for x, the second of the above equations e x = 1/2 Use inverse property (9) to rewrite the above in logarithm form (or take logarithms of both sides) x = ln(1/2) = - ln(2) check Check first solution: x = 0 Evaluate Left Side of equation 2 e 0 + e 0 = 2*1 + 1 =3 Right Side of Equation = 3 Check second solution: x = - ln(2) Evaluate Left Side of equation: 2 e - ln(2) + e -(-ln(2)) = 2/e ln(2) + e ln(2) = = 2/2 + 2 = 3 Right Side of Equation = 3 Conclusion The solutions to the given equation are: x = 0 and x = - ln(2).
Example 7
ln (x + 1) + ln (x) = ln (2) Solution to Example 7Use property (1) above to group the two terms on the left side of the equation ln[ (x + 1) x ] = ln(2) Use property (6) to write the algebraic equation (x + 1) x = 2 Expand and write in standard form x 2 + x - 2 = 0 Solve the above quadratic equation for x to find the solution x = 1 and x = - 2 check Check first solution: x = 1 Left side: ln (1 + 1) + ln (1) = ln (2) , left side is equal to right side ln(2), x = 1 is a solution Check second solution: x = - 2 left side: ln (-2 + 1) + ln (1) = ln ( -1) + ln(1) , ln(-1) is not defined in the real numbers, hence , x = -2 is NOT a solution to the given equation. Conclusion The given equation has one solution: x = 1
Example 8
ln (x + 1) - ln (x) = 2 Solution to Example 8Use property (2) from right to left to group the term on the left side of the equation ln( (x + 1) / x) = 2 Use property (9) to rewrite the above logarithmic equation as follows: (x + 1) / x = e 2 Cross multiply x + 1 = e 2 x Group terms with x on the left and the constant terms on the right x - e 2 x = - 1 factor x out x(1 - e 2) = - 1 Solve for x-fast x = 1 / (e 2 - 1) Check the solution found as an exercise.
Example 9
e 2x e 3x - 3 = 2 Solution to Example 9 Use property (7) to group the exponential terms on the left e 2 x + 3 x - 3 = 2 Simplify and rewrite the equation as follows e 5 x = 5 Use property (9) to rewrite the above exponential equation as follows: 5 x = ln (5) x = ln(5) / 5
Example 10
ln (x 4) + ln (x 2) - ln (x 3) - 2 = 7 Solution to Example 10Use properties (1) and (2) to rewrite the left side as ln (x 4 x 2 / x 3) - 2 = 7 Simplify the expression inside th ln add 2 to both sides of the equation and simplify ln (x 3) = 9 Use property (9) to rewrite the equation as follows x 3 = e9 Solve for x x = ∛(e9) = (e9)1/3 = e3 More References and LinksRules of Logarithms and ExponentialsConvert Logarithms and Exponentials |