Question

If 2to the power x =3to the power y. Prove that (x+3y) z=xy.​

Answer 1

Answer:

2^x  = 3^y  = 6^z

2^x = 6^z

Raise both sides to power y.

   2^{xy} = 6^{zy}    --- (1)

3^y = 6^z

Raise both sides to power x

  3^{xy} = 6^{xz}    --- (2)

multiply (1) and (2)

  =>  (2 * 3)^{x y}  =  6^{z y + x z} = 6^{z(x+y)}

  =>  6^{x y} = 6^{z(x+y)}

  =>  x y =  z(x+y)

  =>  z = (x y) / (x + y)

Answer 2

SOLUTION

GIVEN

$\displaystyle\sf{ {2}^{x} = {3}^{y} = {24}^{z} }$

TO PROVE

$\sf \: x(x +3y)z = xy$

EVALUATION

Here it is given that

$\displaystyle\sf{ {2}^{x} = {3}^{y} = {24}^{z} }$

Let us consider

$\displaystyle\sf{ {2}^{x} = {3}^{y} = {24}^{z} } = k$

Then we have

$\displaystyle\sf{ {k}^{ \frac{1}{x} } = 2 }$

$\displaystyle\sf{ {k}^{ \frac{1}{y} } = 3 }$

$\displaystyle\sf{ {k}^{ \frac{1}{z} } = 24 }$

Now we have

$\sf \: {2}^{3} \times 3 = 24$

$\displaystyle\sf{ \implies \: {k}^{ \frac{3}{x} } \times {k}^{ \frac{1}{y} }={k}^{ \frac{1}{z} } }$

$\displaystyle\sf{ \implies \: {k}^{ \frac{3}{x} + \frac{1}{y} } ={k}^{ \frac{1}{z} } }$

$\displaystyle\sf{ \implies \: \frac{3}{x} + \frac{1}{y} = \frac{1}{z} }$

$\displaystyle\sf{ \implies \: (x + 3y)z = xy }$

Hence proved

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