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

$log {x}^{18}=log {y}^{21}$

$18 log x= 21 log y$

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

$3 \dfrac{log x}{log y} =\dfrac{7}{2}$

$\bold{ 3 log_y x =\dfrac{7}{2}}$

Now,

${x}^{18}={z}^{28}$

By taking LOG on both sides

$log {x}^{18}=log {z}^{28}$

$18 log x =28 log z$

$9 log x =7 \times 2 log z$

$7\dfrac{log z}{log x} = \dfrac{9}{2}$

$\bold{7 log_x z= \dfrac{9}{2}}$

Now,

${y}^{21}={z}^{28}$

By taking LOG on both sides

$log {y}^{21}=log {z}^{28}$

$21 log y=28 log z$

$3 log y = 4 log z$

$3 \dfrac{ log y}{logz }= 4$

$\bold{ 3 log_z y = 4 }$

According to the question:-

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

we have,

$\bold{ 3 log_y x =\dfrac{7}{2}}$

$\bold{7 log_x z= \dfrac{9}{2}}$

$\bold{ 3 log_z y = 4 }$

The series is,

$3 , \: \dfrac{7}{2 } , \: 4 , \: \dfrac{9}{2}$

Here,

$a_1 = 3$

$a_2 = \dfrac{7}{2 }$

$a_3 = 4$

$a_2 = \dfrac{7}{2 }$

Let, us find common difference

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

$d \implies 2 a_2 = a_3+a_1$

$d \implies 2 \times \dfrac{7}{2}= 7$

$d \implies 7 = 7$

Since, common difference is equal!

The series is in AP.