2. Use the table in Question 1 to solve \( x \) : \( 2.1 \quad 2^{x}=4 \) 2.2 \( 2.4 \quad 2^{x}=32 \) 2.5 2.7 \( 3^{x}=\frac{1}{8} \)

Solve the equation by following steps: - step0: Solve for \(x\): \(3^{x}=\frac{1}{8}\) - step1: Take the logarithm of both sides: \(\log_{3}{\left(3^{x}\right)}=\log_{3}{\left(\frac{1}{8}\right)}\) - step2: Evaluate the logarithm: \(x=\log_{3}{\left(\frac{1}{8}\right)}\) - step3: Simplify: \(x=-3\log_{3}{\left(2\right)}\) Solve the equation \( 2^{x}=4 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2^{x}=4\) - step1: Rewrite in exponential form: \(2^{x}=2^{2}\) - step2: Set the exponents equal: \(x=2\) Solve the equation \( 2^{x}=32 \). Solve the equation by following steps: - step0: Solve for \(x\): \(2^{x}=32\) - step1: Rewrite in exponential form: \(2^{x}=2^{5}\) - step2: Set the exponents equal: \(x=5\) Let's solve each equation step by step: ### 2.1 \( 2^{x} = 4 \) To solve this equation, we recognize that \( 4 \) can be expressed as a power of \( 2 \): \[ 4 = 2^2 \] Thus, we can rewrite the equation as: \[ 2^{x} = 2^{2} \] Since the bases are the same, we can set the exponents equal to each other: \[ x = 2 \] ### 2.2 \( 2^{x} = 32 \) Similarly, we express \( 32 \) as a power of \( 2 \): \[ 32 = 2^5 \] Rewriting the equation gives us: \[ 2^{x} = 2^{5} \] Setting the exponents equal: \[ x = 5 \] ### 2.3 \( 3^{x} = \frac{1}{8} \) To solve this equation, we first express \( \frac{1}{8} \) as a power of \( 2 \): \[ \frac{1}{8} = 2^{-3} \] However, since the base on the left side is \( 3 \), we can take the logarithm of both sides: \[ x \log(3) = \log\left(\frac{1}{8}\right) \] Using the property of logarithms: \[ \log\left(\frac{1}{8}\right) = -3 \log(2) \] Thus, we have: \[ x \log(3) = -3 \log(2) \] Solving for \( x \): \[ x = -\frac{3 \log(2)}{\log(3)} \] This can also be expressed as: \[ x = -3 \log_{3}(2) \] ### Summary of Solutions - For \( 2.1 \): \( x = 2 \) - For \( 2.2 \): \( x = 5 \) - For \( 2.3 \): \( x = -3 \log_{3}(2) \)