Problem 37 - Algebra and Trigonometry

Question

Based on material learned earlier in the course. The purpose of these problems is to keep the material fresh in your mind so that you are better prepared for the final exam. Solve: \(\log _{2} \sqrt{x+5}=4\)

Step-by-Step Solution

Answer

x = 251

Step 1 - Understand the Logarithmic Equation

The given equation is \(\log_{2} \sqrt{x+5} = 4\). Notice that it involves a logarithm and a square root term.

Step 2 - Rewrite the Logarithmic Expression

Recall that \(\log_{2}\) represents the power to which we must raise 2 to get the given value. Thus, \(\log_{2} \sqrt{x+5} = 4\) can be rewritten as: \[2^{4} = \sqrt{x+5}\].

Step 3 - Isolate the Radical Expression

Simplify \(2^{4} = 16\) to obtain the equation: \[16 = \sqrt{x+5}\].

Step 4 - Remove the Square Root

Square both sides to eliminate the square root. This gives: \(16^{2} = x + 5\).

Step 5 - Simplify the Equation

Calculate \(16^2\): \(16^2 = 256\). Therefore, the equation becomes: \[256 = x + 5\].

Step 6 - Solve for x

Subtract 5 from both sides to solve for \(x\): \[256 - 5 = x\]. Thus, \(x = 251\).

Key Concepts

logarithms

Logarithms are a way to express exponents. If you have a logarithmic expression like \(\log_{2} y = 4\), this means you are looking for the power to which you need to raise 2 to get \y\. In simpler terms, \2^{4} = y\.
In our exercise, we had \(\log_{2} \sqrt{x+5} = 4\). This means we need to find out what number we have to raise 2 to the power of 4 to get \sqrt{x+5}\. Rewriting the expression helped convert it into a simpler form we could solve step by step.

solving equations

Solving equations involves finding the value of the variable that makes the equation true. Steps to solve equations often include isolating the variable on one side of the equation.
In our case: \(\log_{2} \sqrt{x+5} = 4\) was rewritten to \2^4 = \sqrt{x+5}\. By solving for the square root term, we then simplified and isolated \x\.
It's crucial to perform each operation carefully. For example, squaring both sides of the equation effectively removed the square root and enabled us to determine the precise value of \x\.

square roots

A square root of a number is a value that, when multiplied by itself, gives the original number. For instance, the square root of 16 is 4 because \4 \times 4 = 16\.
In our exercise, we encountered \(\sqrt{x+5}\). To remove the square root, we squared both sides of the equation: \( \sqrt{x+5} = 16 \implies (\sqrt{x+5})^2 = 16^2\).
By squaring both sides, we simplified the equation to \(x + 5 = 256\), paving the way to solve for \x\ directly by subtracting 5 from both sides.

exponential functions

Exponential functions involve expressions with a constant base raised to a variable exponent. They can grow rapidly and are often found in math involving growth and decay.
In the exercise, the equation \(2^4 = \sqrt{x+5}\) represents an exponential function where 2 is the base and 4 is the exponent.
Understanding how exponential functions convert into logarithmic form and back again is vital. Here, we turned an exponential form \(2^4\) into a value, then used that value to further simplify and eventually solve for \x\.