Solution of the system of equation $2 x^{4} = y^{4} + z^{4}, x y z = 8$ knowing that the logarithms $l o g_{y} x, l o g_{z} y, l o g_{x} z$ form a geometric progression is-

x = y = z = 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x = y = 2, z = 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = 2, y = z = 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = y = z = 3

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $x = y = z = 2$
$∵ l o g_{y} x, l o g_{z} y, l o g_{x} z$ are in G.P.
$∴ (l o g_{z} y)^{2} = l o g_{y} x l o g_{x} z$
or $(l o g_{z} y)^{2} = \frac{1}{l o g_{z} y}$
or $(l o g_{z} y)^{3} = 1 \Rightarrow l o g_{z} y = 1$
It is possible when $y = z (x, y, z > 0)$
from $2 x^{4} = y^{4} + z^{4}$
$2 x^{4} = 2 y^{4}$
$x = y = z$
from $x y z = 8$
$x^{3} = 8$
$∴ x = 2 \Rightarrow x = y = z = 2$

Suggest Corrections

Similar questions

{2 x}^{4} = y^{4} + z^{4}

x y z = 8

{l o g}_{y} x, {l o g}_{z} y, {l o g}_{x} z

G . P .

x, y, z

R

x, y, z

(x^{2} + x y + y^{2}) (y^{2} + y z + z^{2}) (z^{2} + z x + x^{2}) = x y z

(x^{4} + x^{2} y^{2} + y^{4}) (y^{4} + y^{2} z^{2} + z^{4}) (z^{4} + z^{2} x^{2} + x^{4}) = x^{3} y^{3} z^{3}

Factors of $27 x^{3} + y^{3} + z^{3} - 9 x y z$ are:

∣ ∣ ∣ ∣ ∣ \begin{matrix} x & y & z x^{2} & y^{2} & z^{2} x^{3} & y^{3} & z^{3} \end{matrix} ∣ ∣ ∣ ∣ ∣ = x y z (x - y) (y - z) (z - x)

x^{3} + y^{3} + z^{3} - 3 x y z = \frac{1}{2} (x + y + z) [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}]

Join BYJU'S Learning Program