Problem 29 - College Algebra

Question

Solve each logarithmic equation in Exercises \(27-44 .\) Be sure to reject any value of \(x\) that produces the logarithm of a negative number or the logarithm of \(0 .\) $$\log _{4}(x+5)=3$$

Step-by-Step Solution

Answer

The solution to the equation \( \log_{4}(x+5)=3 \) is \( x = 59 \)

Step 1: Convert the logarithmic equation into exponential form

Since the equation is \( \log_{4}(x+5)=3 \), we can rewrite it in exponential form: \( 4^{3}=x+5\)

Step 2: Simplify the equation

We can simplify \( 4^{3} \) to get 64. So the equation becomes \( 64 = x + 5 \)

Step 3: Solve for x

To isolate x, we subtract 5 from both sides of the equation which gives us \( x = 64 - 5 = 59 \)

Step 4: Check the solution

We need to check that our solution does not produce the log of a negative number or zero. We input \( x = 59 \) back into our original equation and see that \( \log_{4}(59+5) = \log_{4}(64) \), indeed equals 3, which is a valid solution.

Key Concepts

Exponential Form

When working with logarithmic equations, it's important to understand the concept of exponential form. Logarithms and exponents are closely related, and converting a logarithmic equation into exponential form is often a critical step in solving these equations. For instance, given the logarithmic equation \( \log_{b}(a) = c \), it can be rewritten in exponential form as \( b^{c} = a \). This conversion helps by expressing the original logarithmic statement in a more straightforward manner, making it easier to manipulate algebraically. In our example, the equation \( \log_{4}(x+5)=3 \) becomes \( 4^{3} = x+5 \), effectively transforming a logarithmic relationship into an exponential one.

Solve for x

Once a logarithmic equation is converted into exponential form, the next step is to solve for \( x \). This typically involves isolating \( x \) on one side of the equation. Let's break this down using the equation \( 4^{3} = x+5 \).

• Calculate \( 4^{3} \) to simplify the expression, which gives 64
• Reformulate the equation as \( 64 = x + 5 \)
• Subtract 5 from both sides to solve for \( x \), which gives \( x = 64 - 5 = 59 \)

This isolation step is crucial in identifying the value of \( x \), hence unraveling the solution to the original logarithmic equation.

Checking Solutions

After finding a value for \( x \), verifying this solution is essential to ensure it satisfies the original equation. This involves substituting the solution back into the original logarithmic equation and confirming its validity. For the equation \( \log_{4}(x+5) = 3 \), we substitute \( x = 59 \) to check:

• Calculate \( \log_{4}(59+5) \) which simplifies to \( \log_{4}(64) \)
• Since \( 64 \) is \( 4^3 \), the log base 4 of 64 equals 3

This confirms that our solution \( x = 59 \) correctly satisfies the original logarithmic equation, thus verifying its correctness.

Negative Number Constraint

It's crucial to remember the constraints related to logarithms involving negative numbers or zero. In mathematics, the base of a logarithm (in this case 4) applied to a number must be positive, i.e., \( \log_{b}(a) \) is only defined for \( a > 0 \). When solving logarithmic equations, any calculated solution for \( x \) must be checked to ensure that it does not turn the argument of the logarithm (\( x+5 \) in this case) negative or zero. For instance:

• In \( \log_{4}(x+5) \), the constraint is \( x+5 > 0 \)
• This simplifies to \( x > -5 \)
• The solution \( x = 59 \) comfortably satisfies this constraint

Ensuring that the solution does not violate these constraints is key in establishing the validity of your solution in logarithmic equations.