{x}^(18)={y}^(21)={z}^(28) 3, 3 log_y x, 3 log_z y and 7 log_x z are in?​

Question

`{x}^(18)={y}^(21)={z}^(28)`

3,
`3 log_y x`,
`3 log_z y` and
`7 log_x z` are in?​

Answer 1

`\large\bold{\underline{\underline{Question:-}}}`

`{x}^(18)={y}^(21)={z}^(28)`

3,
`3 log_y x`,
`3 log_z y` and
`7 log_x z` are in??

`\large\bold{\underline{\underline{Answer:-}}}`

The Series in AP.

`\large\bold{\underline{\underline{Explanation:-}}}`

`{x}^(18)={y}^(21)={z}^(28)`

`{x}^(18)={y}^(21)`

By taking LOG on both sides

`log {x}^(18)=log {y}^(21)`

`18 log x= 21 log y`

`2 \ times 3 log x= 7 log y`

`3 (log x)/(log y) =(7)/(2)`

`\bold{ 3 log_y x =(7)/(2)}`

Now,

`{x}^(18)={z}^(28)`

By taking LOG on both sides

`log {x}^(18)=log {z}^(28)`

`18 log x =28 log z`

`9 log x =7 * 2 log z`

`7(log z)/(log x) = (9)/(2)`

`\bold{7 log_x z= (9)/(2)}`

Now,

`{y}^(21)={z}^(28)`

By taking LOG on both sides

`log {y}^(21)=log {z}^(28)`

`21 log y=28 log z`

`3 log y = 4 log z`

`3 ( log y)/(logz )= 4`

`\bold{ 3 log_z y = 4 }`

According to the question:-

3,
`3 log_y x`,
`3 log_z y` and
`7 log_x z` are ...in??

we have,

`\bold{ 3 log_y x =(7)/(2)}`

`\bold{7 log_x z= (9)/(2)}`

`\bold{ 3 log_z y = 4 }`

The series is,

`3 , \: (7)/(2 ) , \: 4 , \: (9)/(2)`

Here,

`a_1 = 3`

`a_2 = (7)/(2 )`

`a_3 = 4`

`a_2 = (7)/(2 )`

Let, us find common difference

`d \implies a_2 - a_1 = = a_3-a_2`

`d \implies 2 a_2 = a_3+a_1`

`d \implies 2 * (7)/(2)= 7`

`d \implies 7 = 7`

Since, common difference is equal!

The series is in AP.