Question

Los números positivos

x,

y,

y

z

satisfacen

xyz = 10^{81}

y

(\log_{10} x)(\log_{10} yz)

+ (\log_{10} y)(\log_{10} z) = 468.

Halla

\small \sqrt{(\log_{10} x)^2 + (\log_{10} y)^2 + (\log_{10} z)^2}.

Positive numbers

x,

y,

and

z

satisfy

xyz = 10^{81}

and

(\log_{10} x)(\log_{10} yz)

+ (\log_{10} y)(\log_{10} z) = 468.

Find

\small \sqrt{(\log_{10} x)^2 + (\log_{10} y)^2 + (\log_{10} z)^2}.

Respuesta

Sea

a = \log_{10} x,

b = \log_{10} y,

c = \log_{10} z:

entonces

a + b + c = 81

y la segunda condición dice

ab + ac + bc = 468.

Let

a = \log_{10} x,

b = \log_{10} y,

c = \log_{10} z:

then

a + b + c = 81

and the second condition says

ab + ac + bc = 468

Usa

(a + b + c)^2

= a^2 + b^2 + c^2

+ 2(ab + ac + bc).

Use

(a + b + c)^2

= a^2 + b^2 + c^2

+ 2(ab + ac + bc)

Sea

a = \log_{10} x,

b = \log_{10} y,

y

c = \log_{10} z.

Tomando logaritmos en

xyz = 10^{81}

se obtiene

a + b + c = 81.

Como

\log_{10} yz = b + c,

la segunda condición es

a(b + c) + bc = ab + ac + bc

= 468.

Elevando al cuadrado la suma,

\begin{aligned} a^2 + b^2 + c^2 &= (a + b + c)^2 \\ &\quad {}- 2(ab + ac + bc) \\ &= 81^2 - 2 \cdot 468 \\ &= 6561 - 936 \\ &= 5625, \end{aligned}

así que el valor pedido es

\sqrt{5625} = 75.

Let

a = \log_{10} x,

b = \log_{10} y,

and

c = \log_{10} z.

Taking logs of

xyz = 10^{81}

gives

a + b + c = 81.

Since

\log_{10} yz = b + c,

the second condition is

a(b + c) + bc = ab + ac + bc

= 468.

Squaring the sum,

\begin{aligned} a^2 + b^2 + c^2 &= (a + b + c)^2 \\ &\quad {}- 2(ab + ac + bc) \\ &= 81^2 - 2 \cdot 468 \\ &= 6561 - 936 \\ &= 5625, \end{aligned}

so the requested value is

\sqrt{5625} = 75.