Problem 60 - Algebra and Trigonometry

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. $$ \log _{2}(4 x+1)=5 $$

Step-by-Step Solution

Answer

The solution to the equation is \(x = 7.75\).

Step 1: Rewrite the Equation in Exponential Form

First, rewrite the logarithmic equation into an equivalent exponential equation. This is done by applying the formula \(y = \log_b x \leftrightarrow b^{y} = x \). So in our case, the equation \(\log _{2}(4x + 1) = 5 \) transforms into \(2^5 = 4x + 1\).

Step 2: Solve Equation

Now that we have the equation in simpler form, we can proceed into solving for \(x\). Start with substracting 1 from both sides: \(32 - 1 = 4x\). Then divide by 4 on both sides to isolate \(x\): \((32 - 1)/4 = x\).

Step 3: Check the Domain

Because logarithms are only defined for positive numbers, any solution that produces a negative result inside the logarithm should be rejected. However, in this case, any real number would satisfy \(4x + 1 > 0\), so there are no domain constraints.

Step 4: Approximate Decimal Value

Finally, calculate the decimal value of \(x\) maintaining two decimal places, if the exact answer is not a whole number. This is done by performing the division from Step 2: \((32 - 1)/4 = 7.75\).

Key Concepts

Exponential Form

The process of converting a logarithmic equation into an exponential equation is called rewriting in exponential form. This conversion is crucial for simplifying and solving logarithmic equations. The basic idea is based on the formula:

• If given a logarithmic equation: \[y = \log_b x\],
• It can be rewritten as: \[b^{y} = x\].

In this context, the logarithmic equation is \(\log _{2}(4x + 1) = 5\). By applying our formula, the equation is transformed into the exponential form \(2^5 = 4x + 1\). This step makes the equation easier to handle.One now simply equates the base (2) raised to the power (5) to the expression \(4x + 1\). Solving this equation leads us to the value of \(x\). Once in exponential form, calculations become more straightforward, facilitating easier solutions.

Domain of Logarithmic Functions

Understanding the domain is critical when working with logarithmic functions. The domain of a logarithmic function involves the set of all possible values that \(x\) can take such that the logarithmic expression is defined.In the context of this problem, where the equation is \(\log _{2}(4x + 1) = 5\), the expression inside the logarithm \(4x + 1\) must be greater than zero. This is because the logarithmic function is only defined for positive arguments.

• So, we need: \(4x + 1 > 0\).

Solving this inequality gives:

• \(4x > -1\)
• \(x > -1/4\)

In conclusion, the domain restriction implies that any real number greater than \(-1/4\) satisfies the domain requirement. For this particular equation, the solution \(x = 7.75\) falls within the permissible domain, so no values are rejected.

Decimal Approximation

Sometimes, even if we have solved the equation exactly, it is necessary to express the solution in decimal form for practical purposes or when dealing with complex numbers. A decimal approximation provides a numerical estimate. In our example, once the exact value of \(x\) is determined as \((32 - 1)/4\), a calculator is typically used to carry out the division:

• \((31)/4 = 7.75\).

This gives us a numerical solution accurate to two decimal places.It's important to round to the specified number of decimal places to avoid potential errors in interpretations or further calculations.This decimal approximation can make the solution more tangible and help in scenarios where an estimate is more useful than an exact number.

Exact Solution

Achieving the exact solution means finding a value for \(x\) that satisfies the original equation without resorting to an approximation. This is crucial in precise mathematical analysis or in settings, for instance, where the solution is part of a larger symbolic expression. To find the exact solution of the initial problem, we converted the logarithm to exponential form, arriving at:

• \(2^5 = 4x + 1\).

Carrying out the operations, we began by subtracting 1:

• \(32 - 1 = 4x\)

This led to:

• \(31 = 4x\)

Finally, dividing both sides by 4 isolates \(x\):

• \(x = \frac{31}{4}\)
• Resulting in the exact solution: \(x = 7.75\)

The power of exact solutions comes from their complete accuracy in any context, ensuring no details are lost in translation.