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

Solución:

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 $\sqrt{5625} = 75.$

2010 AIME II Problema 5

Solución:

El Problema 5 en otros años