Problem 56 - 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. $$ \log _{2}(x+50)=5 $$

Step-by-Step Solution

Answer

The exact solution of the logarithmic equation is \(x = -18\)

Step 1: Transform Logarithm to Exponential form

The first step is to convert the logarithmic equation to exponential form. Note how the base of the logarithm \(2\), the argument of the logarithm \(x + 50\) and the value \(5\) relate in the equation. Writing this as an equivalent exponential equation gives: \(2^5 = x + 50\).

Step 2: Solve for x

The next step involves isolating \(x\). This is done by subtracting \(50\) from each side of the equation, which yields \(x = 2^5 - 50\).

Step 3: Calculate Exact Answer

The exact answer can be found by calculating \(2^5\) which equals to \(32\), and then subtract \(50\), which gives an answer of \(x = -18\).

Step 4: Check Solution

The final step is to verify that the answer is within the domain of the original logarithmic expression. Note that the input to a logarithm (x + 50 in this case) must be greater than \(0\), which implies \(x > -50\). Since our solution \(-18\) is greater than \(-50\). We don't reject it.

Key Concepts

Logarithm to Exponential Form

Understanding the relationship between logarithms and exponentials is crucial when solving logarithmic equations. At its core, a logarithmic equation like \(\log _{b}(x)=y\) can be rewritten in exponential form as \(b^y = x\). This transformation is fundamental because it leverages our familiarity with exponentiation to solve for the unknown variable.

To convert \(\log _{2}(x+50)=5\) to exponential form, consider the base \(2\), the exponent \(5\), and the result \(x+50\). Applying the conversion yields \(2^5 = x + 50\), enabling us to use arithmetic operations to isolate and solve for \(x\). This step is the bridge between two important mathematical concepts, providing a clearer path to the solution.

Isolating the Variable

Isolating the variable in an equation is a key step in solving for that variable. The goal is to manipulate the equation so that the variable stands alone on one side, making it easy to identify the solution. After converting a logarithmic equation to exponential form, the next step is to isolate the variable.

For the equation \(2^5 = x + 50\), subtracting \(50\) from both sides allows us to isolate \(x\) on one side of the equation. This straightforward action, \(x = 2^5 - 50\), brings us closer to the answer by removing addition or subtraction operations that surround the variable. Performing the calculation results in the exact value for \(x\). Students should always simplify to the most reduced form to find the precise solution.

Domain of Logarithmic Functions

The domain of logarithmic functions is a critical concept, as it specifies the range of acceptable inputs to the logarithm. For any logarithmic function \(\log_b(x)\), the argument \(x\) must be greater than zero since logging negative numbers or zero is undefined in the realm of real numbers.

When solving logarithmic equations, always ensure your solution is within the domain. In our case, the solution must satisfy \(x + 50 > 0\) since the domain restriction comes from the argument of \(\log_2(x + 50)\). Consequently, \(x > -50\). If we find a solution outside of this range, we must reject it, as it does not belong to the function's domain. However, for the equation \(\log _{2}(x+50)=5\), the calculated answer of \(x = -18\) is within the permissible range, making it a valid solution.