Suppose, $$\log_3 x = \log_{12} y = a$$, where $$x, y$$ are positive numbers. If $$G$$ is the geometric mean o

Question

Suppose, $$\log_3 x = \log_{12} y = a$$, where $$x, y$$ are positive numbers. If $$G$$ is the geometric mean of x and y, and $$\log_6 G$$ is equal to

Accepted Answer

a. We know that log ⁡ 3 x = a \log_3 x = a lo g 3 ​ x = a and log ⁡ 12 y = a \log_{12} y=a lo g 12 ​ y = a Hence, x = 3 a x = 3^a x = 3 a and y = 12 a y=12^a y = 1 2 a Therefore, the geometric mean of x x x and y y y equals x × y \sqrt{x 	imes y} x × y ​ This equals 3 a × 12 a = 6 a \sqrt{3^a 	imes 1

Suppose, $\log_3 x = \log_{12} y = a$ , where $x, y$ are positive numbers. If $G$ is the geometric mean of x and y, and $\log_6 G$ is equal to

Suppose, $\log_3 x = \log_{12} y = a$ , where $x, y$ are positive numbers. If $G$ is the geometric mean of x and y, and $\log_6 G$ is equal to

Suppose, log⁡3x=log⁡12y=a\log_3 x = \log_{12} y = alog3​x=log12​y=a, where x,yx, yx,y are positive numbers. If GGG is the geometric mean of x and y, and log⁡6G\log_6 Glog6​G is equal to

Suppose, log⁡3x=log⁡12y=a\log_3 x = \log_{12} y = alog3​x=log12​y=a, where x,yx, yx,y are positive numbers. If GGG is the geometric mean of x and y, and log⁡6G\log_6 Glog6​G is equal to

Suppose, $\log_3 x = \log_{12} y = a$ , where $x, y$ are positive numbers. If $G$ is the geometric mean of x and y, and $\log_6 G$ is equal to

Suppose, $\log_3 x = \log_{12} y = a$ , where $x, y$ are positive numbers. If $G$ is the geometric mean of x and y, and $\log_6 G$ is equal to