Question

what is the solution to the equation below?
$3\log_{4} x = \log_{4} 32 + \log_{4} 2$
$\bigcirc$ $x = -8$
$\bigcirc$ $x = -4$
$\bigcirc$ $x = 4$
$\bigcirc$ $x = 8$

Explanation:

Step1: Use log addition rule

The logarithm addition rule states that \(\log_b a + \log_b c=\log_b(ac)\). So, \(\log_4 32+\log_4 2=\log_4(32\times2)=\log_4 64\).

Step2: Simplify \(\log_4 64\)

We know that \(4^3 = 64\), so \(\log_4 64 = 3\). Now the equation becomes \(3\log_4 x=3\).

Step3: Divide both sides by 3

Divide each side of \(3\log_4 x = 3\) by 3, we get \(\log_4 x=1\).

Step4: Convert to exponential form

The logarithmic equation \(\log_b a = c\) is equivalent to \(b^c=a\). So, \(\log_4 x = 1\) implies \(4^1=x\), so \(x = 4\)? Wait, no, wait. Wait, let's check again. Wait, \(4^3=64\), so \(\log_4 64 = 3\), so \(3\log_4 x=3\), divide by 3: \(\log_4 x = 1\)? Wait, no, wait \(32\times2 = 64\), \(4^3=64\), so \(\log_4 64 = 3\), so equation is \(3\log_4 x=3\), so \(\log_4 x = 1\)? Wait, no, \(3\log_4 x=\log_4 x^3\) (power rule of logs: \(n\log_b a=\log_b a^n\)). So original equation: \(\log_4 x^3=\log_4(32\times2)=\log_4 64\). Since the logs have the same base and are equal, their arguments are equal: \(x^3=64\). Then \(x=\sqrt[3]{64}=4\)? Wait, no, \(\sqrt[3]{64} = 4\)? Wait, \(4^3=64\), yes. Wait, but let's check the options. Wait, the options have \(x = 4\) and \(x = 8\). Wait, maybe I made a mistake. Wait, \(32\times2=64\), \(\log_4 64\): \(4^3 = 64\), so \(\log_4 64 = 3\). So equation: \(3\log_4 x=3\), so \(\log_4 x = 1\), so \(x = 4^1=4\). But wait, let's check with \(x = 4\): left side \(3\log_4 4=3\times1 = 3\), right side \(\log_4 32+\log_4 2=\log_4(32\times2)=\log_4 64 = 3\). So it works. But wait, the option \(x = 4\) is there. Wait, but maybe I messed up the power rule. Wait, \(3\log_4 x=\log_4 x^3\), so \(\log_4 x^3=\log_4 64\), so \(x^3=64\), so \(x = 4\), since \(4^3=64\). So the answer is \(x = 4\). Wait, but let's check the other way. Wait, maybe I made a mistake in the addition. \(32 + 2\) is not, it's multiplication. Yes, log addition is multiplication of arguments. So that's correct. So \(x = 4\) is the solution. Wait, but the option \(x = 4\) is present. So the correct answer is \(x = 4\).

Wait, no, wait, wait. Wait, \(3\log_4 x=\log_4 x^3\), and \(\log_4 32+\log_4 2=\log_4(32\times2)=\log_4 64\). So \(x^3=64\), so \(x = \sqrt[3]{64}=4\). Yes, that's correct. So the solution is \(x = 4\).

Answer:

C. \(x = 4\) (assuming the options are labeled as A. \(x=-8\), B. \(x = - 4\), C. \(x = 4\), D. \(x = 8\))