what is the solution to the equation below?\n$log_{6}4x^{2}-log_{6}x = 2$\n$x=\frac{1}{12}$\n$x=\frac{3}{2}$\…

what is the solution to the equation below?\n$log_{6}4x^{2}-log_{6}x = 2$\n$x=\frac{1}{12}$\n$x=\frac{3}{2}$\n$x = 3$\n$x = 9$

what is the solution to the equation below?\n$log_{6}4x^{2}-log_{6}x = 2$\n$x=\frac{1}{12}$\n$x=\frac{3}{2}$\n$x = 3$\n$x = 9$

Answer

Explanation:

Step1: Use log - rule

Use the quotient - rule of logarithms $\log_aM-\log_aN = \log_a\frac{M}{N}$. So, $\log_64x^{2}-\log_6x=\log_6\frac{4x^{2}}{x}=\log_64x$. The equation becomes $\log_64x = 2$.

Step2: Convert to exponential form

By the definition of logarithms, if $\log_ab = c$, then $b=a^{c}$. Here, $a = 6$, $b = 4x$, and $c = 2$. So, $4x=6^{2}$.

Step3: Solve for x

Since $6^{2}=36$, the equation $4x = 36$ gives $x=\frac{36}{4}=9$.

Answer:

$x = 9$