5) \( \log _{2} x+\log _{x} 16 \)

To simplify the expression \( \log_{2} x + \log_{x} 16 \), let's break it down step by step. ### Step 1: Express Both Logarithms with the Same Base First, notice that both logarithms can be expressed in terms of base 2: 1. \( \log_{2} x \) is already in base 2. 2. To change the base of \( \log_{x} 16 \) to base 2, use the **Change of Base Formula**: \[ \log_{x} 16 = \frac{\log_{2} 16}{\log_{2} x} \] Since \( 16 = 2^4 \), we have: \[ \log_{2} 16 = 4 \] Therefore: \[ \log_{x} 16 = \frac{4}{\log_{2} x} \] ### Step 2: Combine the Expressions Now, substitute the expression for \( \log_{x} 16 \) back into the original equation: \[ \log_{2} x + \log_{x} 16 = \log_{2} x + \frac{4}{\log_{2} x} \] ### Step 3: Simplify the Expression Let \( y = \log_{2} x \). The expression becomes: \[ y + \frac{4}{y} \] This is a simplified form of the original expression. ### Additional Insight: Minimizing the Expression If you're interested in finding the minimum value of the expression \( y + \frac{4}{y} \), you can use the **Arithmetic Mean-Geometric Mean (AM-GM) Inequality**: \[ y + \frac{4}{y} \geq 2\sqrt{y \cdot \frac{4}{y}} = 2\sqrt{4} = 4 \] Equality occurs when \( y = \frac{4}{y} \), which implies \( y = 2 \). Therefore, the minimum value of the expression is **4** when \( \log_{2} x = 2 \), or equivalently, when \( x = 4 \). ### Final Simplified Form \[ \log_{2} x + \log_{x} 16 = \log_{2} x + \frac{4}{\log_{2} x} \] Or, using \( y = \log_{2} x \): \[ y + \frac{4}{y} \] ### Summary The expression \( \log_{2} x + \log_{x} 16 \) simplifies to \( \log_{2} x + \frac{4}{\log_{2} x} \). Additionally, its minimum value is 4 when \( x = 4 \).