CrazySnowGirL
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[tex]{x}^{18}={y}^{21}={z}^{28}[/tex]
3, [tex]3 log_y x [/tex], [tex]3 log_z y [/tex] and [tex]7 log_x z [/tex] are in?​

Sagot :

[tex]\large\bold{\underline{\underline{Question:-}}}[/tex]

[tex]{x}^{18}={y}^{21}={z}^{28}[/tex]

3, [tex]3 log_y x [/tex], [tex]3 log_z y [/tex] and [tex]7 log_x z [/tex] are in??

[tex]\large\bold{\underline{\underline{Answer:-}}}[/tex]

The Series in AP.

[tex]\large\bold{\underline{\underline{Explanation:-}}}[/tex]

[tex]{x}^{18}={y}^{21}={z}^{28} [/tex]

[tex]{x}^{18}={y}^{21}[/tex]

By taking LOG on both sides

[tex]log {x}^{18}=log {y}^{21} [/tex]

[tex]18 log x= 21 log y[/tex]

[tex]2 \ times 3 log x= 7 log y[/tex]

[tex] 3 \dfrac{log x}{log y} =\dfrac{7}{2}[/tex]

[tex]\bold{ 3 log_y x =\dfrac{7}{2}}[/tex]

Now,

[tex]{x}^{18}={z}^{28} [/tex]

By taking LOG on both sides

[tex]log {x}^{18}=log {z}^{28} [/tex]

[tex]18 log x =28 log z [/tex]

[tex] 9 log x =7 \times 2 log z [/tex]

[tex] 7\dfrac{log z}{log x} = \dfrac{9}{2}[/tex]

[tex]\bold{7 log_x z= \dfrac{9}{2}}[/tex]

Now,

[tex]{y}^{21}={z}^{28} [/tex]

By taking LOG on both sides

[tex]log {y}^{21}=log {z}^{28} [/tex]

[tex]21 log y=28 log z [/tex]

[tex] 3 log y = 4 log z [/tex]

[tex]3 \dfrac{ log y}{logz }= 4 [/tex]

[tex]\bold{ 3 log_z y = 4 }[/tex]

According to the question:-

3, [tex]3 log_y x [/tex], [tex]3 log_z y [/tex] and [tex]7 log_x z [/tex] are ...in??

we have,

[tex]\bold{ 3 log_y x =\dfrac{7}{2}}[/tex]

[tex]\bold{7 log_x z= \dfrac{9}{2}}[/tex]

[tex]\bold{ 3 log_z y = 4 }[/tex]

The series is,

[tex]3 , \: \dfrac{7}{2 } , \: 4 , \: \dfrac{9}{2}[/tex]

Here,

[tex]a_1 = 3 [/tex]

[tex]a_2 = \dfrac{7}{2 } [/tex]

[tex]a_3 = 4 [/tex]

[tex]a_2 = \dfrac{7}{2 } [/tex]

Let, us find common difference

[tex]d \implies a_2 - a_1 = = a_3-a_2 [/tex]

[tex]d \implies 2 a_2 = a_3+a_1 [/tex]

[tex]d \implies 2 \times \dfrac{7}{2}= 7 [/tex]

[tex]d \implies 7 = 7 [/tex]

Since, common difference is equal!

The series is in AP.

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