Problem 78 - College Algebra

Question

Solve each logarithmic equation in Exercises \(49-92 .\) 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. $$ \log (5 x+1)=\log (2 x+3)+\log 2 $$

Step-by-Step Solution

Answer

The solution for the given logarithmic equation is \(x = 5\).

Step 1: Combine the two logarithms on the right side using the product rule

According to the product rule of logarithms, \(log(M) + log(N) = log(MN)\). Let's apply this rule to our equation. Therefore, the equation \(\log (5x+1) = \log (2x+3) + \log 2\) becomes \(\log (5x+1) = \log((2x+3)*2)\) or \(\log (5x+1) = \log (4x+6)\)

Step 2: Remove the logs

Since both sides of the equation is in logarithms, we can remove the logarithms. This gives us \(5x + 1 = 4x + 6\)

Step 3: Solve for \(x\)

If we rearrange the equation, we see that \(5x - 4x = 6 - 1\), which simplifies to \(x = 5\)

Step 4: Check the validity of the solution in the original equation

Plugging \(x = 5\) back into the original equation \(\log(5x + 1) = \log(2x + 3) + log 2\), after calculation, we observe that both sides of the equation are equal. Therefore \(x = 5\) is a valid solution considering the domain of original logarithmic expressions.

Key Concepts

Domain of Logarithmic Expressions

When dealing with logarithmic equations, it is crucial to check the domain of the expressions involved. The domain of a logarithmic function is defined by the requirement that the argument of the logarithm must always be positive. This means that whatever is inside the logarithm, such as \(5x+1\) or \(2x+3\), must be greater than zero.
Applying this to the exercise, we need:

• \(5x + 1 > 0\), which simplifies to \(x > -\frac{1}{5}\)
• \(2x + 3 > 0\), which simplifies to \(x > -\frac{3}{2}\)

The more restrictive condition would be \(x > -\frac{1}{5}\) as it covers the second condition automatically. Therefore, any solution for \(x\) must fulfill this requirement to be valid in the context of the original expression.
By ensuring your solution lives within this domain, you can confidently reject any extraneous solutions that might otherwise seem valid.

Product Rule for Logarithms

The product rule for logarithms is a handy tool when simplifying logarithmic expressions. It states that the logarithm of a product is equal to the sum of the logarithms. Mathematically, \(\log(M) + \log(N) = \log(M \times N)\).
In our exercise, this rule was applied to combine \(\log(2x+3) + \log 2\) into a single term: \(\log((2x+3) \times 2)\) or \(\log(4x+6)\).
By combining the logarithms, we are able to equate the expressions in a way that simplifies the equation remarkably. This is because simplifying the expression into a single logarithm on each side means we can more easily apply properties of equality in subsequent steps.

Validity of Solutions

After solving a logarithmic equation, we must verify the validity of the solution within the context of the problem. For logarithmic functions, this involves checking if the solution lies within the domain we established earlier.
In this case, we arrived at the solution \(x = 5\). Plugging this back into the original equation ensures that both sides equate when substituted:

• \(5x + 1 = 26\) for \(x = 5\)
• \(2x + 3 = 13\) and adding the \(\log 2\)

We see that both sides agree because calculations verify equality. Moreover, since \(x = 5\) is greater than \(-\frac{1}{5}\), it lies within the domain, confirming its validity as a solution that respects all conditions of the logarithmic expressions.

Solving Equations

Solving logarithmic equations involves a process of simplification and verification. Once two sides of the equation are simplified, often by using rules such as the product rule, solutions can be found by equating the arguments of the logarithmic expressions.
In this exercise, once we simplified the equation to \(5x+1 = 4x+6\), the logs could be disregarded. However, this simplification step is permissible only when both components originally expressed logarithmic equality. From there, we can solve traditional algebraically:

• Move the variable terms: \(5x - 4x = 6 - 1\)
• Simplify: \(x = 5\)

This step-by-step process not only provides the solution but ensures that none of the rules or conditions associated with logarithmic functions are violated, preserving the integrity of the mathematical operation.