Problem 76 - College Algebra Essentials

Question

Solve each logarithmic equation. Be sure to reject any value of \(x\) that is not in the domain of the original logarithmic expressions. Give the exact answer. Then, where necessary, use a calculator to obtain a decimal approximation, correct to two decimal places, for the solution. $$\log _{2}(x-3)+\log _{2} x-\log _{2}(x+2)=2$$

Step-by-Step Solution

Answer

Thus, the only valid solution is \(x = 8\).

Step 1: Simplify using Logarithm Properties

The three logarithmic expressions can be simplified into one using log properties. \\[\log _{2}(x-3)+\log _{2} x -\log _{2}(x+2) = \log _{2}\left(\frac{x(x-3)}{x+2}\right)=2\\]

Step 2: Convert the Logarithmic Equation into an Exponential Equation

Afterwards, the equation is transformed from its logarithmic form into its equivalent exponential form. Thus the equation becomes: \\[2^2 = \frac{x(x-3)}{x+2}\\]

Step 3: Solve for x

Cross multiply and simplify this equation to solve for x. \\[4(x+2) = x(x-3) \\4x+8 = x^2-3x\\x^2-7x+8=0\\] This quadratic equation can be factored to get (x-1)(x-8)=0.

Step 4: Find the Values of x

Set each factor equal to zero and solve for x to get: x=1 , x=8.

Step 5: Verify the Domain of the Logarithmic Expression

But we must check these solutions against the original logarithmic expressions to make sure they don't result in a negative number or zero inside any of the logarithms. Doing this, we see that x=1 does indeed result in a negative number inside the first logarithm (1-3 = -2). As such, x=1 is not a valid solution and should be rejected.