Question

question
solve for the exact value of x.
\log_{6}(6x) + 4\log_{6}(2) = 0

Explanation:

Step1: Use logarithm power rule

The power rule of logarithms states that \(n\log_b(a)=\log_b(a^n)\). Apply this to \(4\log_6(2)\):
\(4\log_6(2)=\log_6(2^4)=\log_6(16)\)
So the equation becomes \(\log_6(6x)+\log_6(16) = 0\)

Step2: Use logarithm product rule

The product rule of logarithms states that \(\log_b(a)+\log_b(c)=\log_b(ac)\). Apply this to \(\log_6(6x)+\log_6(16)\):
\(\log_6(6x\times16)=\log_6(96x)=0\)

Step3: Convert logarithmic equation to exponential form

Recall that if \(\log_b(y)=k\), then \(y = b^k\). For \(\log_6(96x)=0\), we have:
\(96x=6^0\)
Since \(6^0 = 1\), the equation is \(96x=1\)

Step4: Solve for x

Divide both sides of the equation \(96x = 1\) by 96:
\(x=\frac{1}{96}\)

Answer:

\(\frac{1}{96}\)