Algebra ExamplesPopular Problems Show Algebra Solve for x log base 5 of x+30=3 Rewrite in exponential form using the definition of a logarithm. If and are positive real numbers and , then is equivalent to . Solve for . Tap for more steps... Rewrite the equation as . Raise to the power of . Move all terms not containing to the right side of the equation. Tap for more steps... Subtract from both sides of the equation. Subtract from . As you well know that, a logarithm is a mathematical operation that is the inverse of exponentiation. The logarithm of a number is abbreviated as “log.” Before we can get into solving logarithmic equations, let’s first familiarize ourselves with the following rules of logarithms:
The product rule states that the sum of two logarithms is equal to the product of the logarithms. The first law is represented as; ⟹ log b (x) + log b (y) = log b (xy)
The difference of two logarithms x and y is equal to the ratio of the logarithms. ⟹ log b (x) – log b (y) = log (x/y)
⟹ log b (x) n = n log b (x)
⟹ log b x = (log a x) / (log a b)
The logarithm of any positive number to the same base of that number is always 1. Example:
How to Solve Logarithmic Equations?An equation containing variables in the exponents is knowns as an exponential equation. In contrast, an equation that involves the logarithm of an expression containing a variable is referred to as a logarithmic equation. The purpose of solving a logarithmic equation is to find the value of the unknown variable. In this article, we will learn how to solve the general two types of logarithmic equations, namely:
How to solve equations with logarithms on one side?Equations with logarithms on one side take log b M = n ⇒ M = b n. To solve this type of equations, here are the steps:
Example 1 Solve log 2 (5x + 7) = 5 Solution Rewrite the equation to exponential form logs 2 (5x + 7) = 5 ⇒ 2 5 = 5x + 7 ⇒ 32 = 5x + 7 ⇒ 5x = 32 – 7 5x = 25 Divide both sides by 5 to get x = 5 Example 2 Solve for x in log (5x -11) = 2 Solution Since the base of this equation is not given, we therefore assume the base of 10. Now change the write the logarithm in exponential form. ⇒ 102 = 5x – 11 ⇒ 100 = 5x -11 111= 5x 111/5 = x Hence, x = 111/5 is the answer. Example 3 Solve log 10 (2x + 1) = 3 Solution Rewrite the equation in exponential form log10 (2x + 1) = 3n⇒ 2x + 1 = 103 ⇒ 2x + 1 = 1000 2x = 999 On dividing both sides by 2, we get; x = 499.5 Verify your answer by substituting it in the original logarithmic equation; ⇒ log10 (2 x 499.5 + 1) = log10 (1000) = 3 since 103 = 1000 Example 4 Evaluate ln (4x -1) = 3 Solution Rewrite the equation in exponential form as; ln (4x -1) = 3 ⇒ 4x – 3 =e3 But as you know, e = 2.718281828 4x – 3 = (2.718281828)3 = 20.085537 x = 5.271384 Example 5 Solve the logarithmic equation log 2 (x +1) – log 2 (x – 4) = 3 Solution First simplify the logarithms by applying the quotient rule as shown below. log 2 (x +1) – log 2 (x – 4) = 3 ⇒ log 2 [(x + 1)/ (x – 4)] = 3 Now, rewrite the equation in exponential form ⇒2 3 = [(x + 1)/ (x – 4)] ⇒ 8 = [(x + 1)/ (x – 4)] Cross multiply the equation ⇒ [(x + 1) = 8(x – 4)] ⇒ x + 1 = 8x -32 7x = 33 …… (Collecting the like terms) x = 33/7 Example 6 Solve for x if log 4 (x) + log 4 (x -12) = 3 Solution Simplify the logarithm by using the product rule as follows; log 4 (x) + log 4 (x -12) = 3 ⇒ log 4 [(x) (x – 12)] = 3 ⇒ log 4 (x2 – 12x) = 3 Convert the equation in exponential form. ⇒ 43 = x2 – 12x ⇒ 64 = x2 – 12x Since this is a quadratic equation, we therefore solve by factoring. x2 -12x – 64 ⇒ (x + 4) (x – 16) = 0 x = -4 or 16 When x = -4 is substituted in the original equation, we get a negative answer which is imaginary. Therefore, 16 is the only acceptable solution. How to solve equations with logarithms on both sides of the equation?The equations with logarithms on both sides of the equal to sign take log M = log N, which is the same as M = N. The procedure of solving equations with logarithms on both sides of the equal sign.
Example 7 Solve log 6 (2x – 4) + log 6 (4) = log 6 (40) Solution First, simplify the logarithms. log 6 (2x – 4) + log 6 (4) = log 6 (40) ⇒ log 6 [4(2x – 4)] = log 6 (40) Now drop the logarithms ⇒ [4(2x – 4)] = (40) ⇒ 8x – 16 = 40 ⇒ 8x = 40 + 16 8x= 56 x = 7 Example 8 Solve the logarithmic equation: log 7 (x – 2) + log 7 (x + 3) = log 7 14 Solution Simplify the equation by applying the product rule. Log 7 [(x – 2) (x + 3)] = log 7 14 Drop the logarithms. ⇒ [(x – 2) (x + 3)] = 14 Distribute the FOIL to get; ⇒ x 2 – x – 6 = 14 ⇒ x 2 – x – 20 = 0 ⇒ (x + 4) (x – 5) = 0 x = -4 or x = 5 when x = -5 and x = 5 are substituted in the original equation, they give a negative and positive argument respectively. Therefor, x = 5 is the only acceptable solution. Example 9 Solve log 3 x + log 3 (x + 3) = log 3 (2x + 6) Solution Given the equation; log 3 (x2 + 3x) = log 3 (2x + 6), drop the logarithms to get; (x – 2) (x + 3) = 0 By verifying both values of x, we get x = 2 to be the correct answer. Example 10 Solve log 5 (30x – 10) – 2 = log 5 (x + 6) Solution log 5 (30x – 10) – 2 = log 5 (x + 6) This equation can be rewritten as; ⇒ log 5 (30x – 10) – log 5 (x + 6) = 2 Simplify the logarithms log 5 [(30x – 10)/ (x + 6)] = 2 Rewrite logarithm in exponential form. ⇒ 52 = [(30x – 10)/ (x + 6)] ⇒ 25 = [(30x – 10)/ (x + 6)] On cross multiplying, we get; ⇒ 30x – 10 = 25 (x + 6) ⇒ 30x – 10 = 25x + 150 ⇒ 30x – 25x = 150 + 10 ⇒ 5x = 160 x = 32 What is a solution equation?The solution of an equation is the set of all values that, when substituted for unknowns, make an equation true. For equations having one unknown, raised to a single power, two fundamental rules of algebra, including the additive property and the multiplicative property, are used to determine its solutions.
How do you solve log5 4?1 Answer. By definition, y=logax⇔ay=x.. ∴log54==y⇔5y=4.. ⇔y=log4log5=0,861.. |