Question

Square abcd has area 36, and abab is parallel to the x axis. Vertices a, b and c are on the graphs of y=logax, y=2 logaxy=logax, y=2 logax and y=3 logax,y=3 logax, respectively. What is a?

Answer 1

Answer:

write.each.of.the.following.as.percent.

$1.8$

Answer 2

Given:

The area of square ABCD is 36

Vertices AB is parallel to the x-axis

vertices A, B, and C are on the graphs

To Find

'a'

Solution:

All the graphs of y = $log{a}$ x, y =  2$log_{a}$ x, and y = 3$log_{a}$ x have the domain (0,∞) and it is farthest to the right.

Since A is on the graph of y= $log_{a}$ x and B is on y= 2$log_{a}$ x,

Vertices AB is parallel to x-axis      (given)

let A be the point (x, y)

then point B and C are (x - 6, y) and (x- 6, y+ 6) respectively.

Substituting, y= $log_{a}$ x, y= 2$log_{a}$ x- 6, and y+ 6= 3$log_{a}$ x- 6.

After rearranging we get, $a^{y}$= x..(1)

                                          $a^{y}$= (x- 6)²..(2)

                                          $a^{y}$+6= (x-6)³..(3)

⇒ x = (x - 6)²                (using equation 1 and 2)

⇒ x = 9, x = -4              

But, as we know a cannot be in negative. So, considering x = 9.

⇒ $a^{y}$+6/$a^{y}$= (x-6)³/(x-6)²

⇒ $a^{6}$ = x - 6       (using equation 2 and 3)

⇒ $a^{6}$ = 9 - 6

⇒ $a^{6}$ = 3

⇒ a = $\sqrt[6]{3}$

Therefore, a is $\sqrt[6]{3}$.