Problem 65 - Introductory and Intermediate Algebra for College Students 4th

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. $$\log _{5} x+\log _{5}(4 x-1)=1$$

Step-by-Step Solution

Answer

To solve this equation, combine the logarithmic expressions, convert the logarithm to an exponent, solve the resulting equation and check if the solutions are valid. The approximate decimal solutions to the equation which fit in the domain of the original equation are the exact answer.

Step 1: Combine the Logarithmic Expressions

Using the logarithmic property \(\log a + \log b = \log (a \cdot b)\), combine \(\log _{5} x\) and \(\log _{5}(4 x-1)\) to get \(\log _{5} (x \cdot (4x - 1))\)

Step 2: Convert the Logarithm to an Exponent

The equation \(\log _{5} (x \cdot (4x - 1)) = 1\) can be written in exponential form as \(5^{1}=x \cdot (4x - 1)\)

Step 3: Solve the Resulting Equation

The equation \(5^{1}=x \cdot (4x - 1)\) simplifies to \(5 = 4x^{2} - x\). This can be rearranged to the standard form of a quadratic equation \(4x^{2} - x - 5 = 0\). Solving the quadratic equation will give two possible values for \(x\)

Step 4: Check if the solutions are in the domain

Finally, check that the obtained solutions for \(x\) are valid in the original equations. Recall that you can only take a logarithm of a positive number, so any solution where \(x\) or \(4x -1\) is less than or equal to 0 must be rejected.