question\nsolve for the exact value of x.\n\\log_{6}(7x) + \\log_{6}(4) = 4

Explanation:

Step1: Apply log addition rule

Using the property $\log_a(b) + \log_a(c) = \log_a(bc)$, we combine the left - hand side: $\log_6(7x)+\log_6(4)=\log_6(7x\times4)=\log_6(28x)$ So the equation becomes $\log_6(28x) = 4$

Step2: Convert to exponential form

Recall that if $\log_a(y)=k$, then $y = a^k$. For our equation $\log_6(28x)=4$, we have: $28x=6^4$

Step3: Calculate $6^4$ and solve for x

We know that $6^4 = 6\times6\times6\times6=1296$. So the equation $28x = 1296$ can be solved for $x$ by dividing both sides by 28: $x=\frac{1296}{28}=\frac{324}{7}$

Answer:

$\frac{324}{7}$