Problem 67 - 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. $$\log _{5} x+\log _{5}(4 x-1)=1$$

Step-by-Step Solution

Answer

The solution for the equation is \(x=0.25\).

Step 1: Combining Logarithmic Expressions

First, use the property of logarithms to simplify the left-hand side of the equation: \(\log _{5} x+\log _{5}(4 x-1) = \log _{5}(x* (4x-1))\). The equation becomes \(\log _{5}(4x^2 - x)=1\).

Step 2: Applying the Definition of Logarithm

A logarithm \(\log_a b = c\) can be converted to the form \(a^c = b\). So apply this to convert the equation to exponential form, leading to \(5^1 = 4x^2 - x\), simplified to \(5 = 4x^2 - x\).

Step 3: Finding Roots for the Quadratic Equation

Now, rearrange the equation to be in the form of a standard quadratic equation: \(4x^2 - x - 5=0\). The solutions of this equation can be found by either factoring or applying the quadratic formula, which is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). If the exercise doesn't specify which method to use, the quadratic formula is a good option as it always works.

Step 4: Applying the Quadratic Formula

Here a=4, b=-1, and c=-5. Substituting these values into the quadratic formula leads to \(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(4)(-5)}}{2(4)} = \frac{1 \pm \sqrt{81}}{8}\). This gives two possible values for \(x\): \(x=0.25\) and \(x=-2.5\).

Step 5: Checking for Valid Logarithmic Domain

The domain of a logarithmic function is \(x > 0\), so check that each potential solution falls in this domain. For the potential solutions, \(x=0.25 > 0\) and \(x=-2.5 < 0\), only \(x=0.25\) fits in this domain, so \(x=-2.5\) would be rejected.

Step 6: Rounding the Solution to Two Decimal Places

As a final step, round the solution \(x=0.25\) to have two decimal places, leaving it as \(x=0.25\) as there is nothing to round, since it already fits these specifications.