Question

Supongamos que $a,$ $b,$ y $c$ son números reales positivos tales que $a^{\log_3 7} = 27,$ $b^{\log_7 11} = 49,$ y $c^{\log_{11} 25} = \sqrt{11}.$ Halla $a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.$

Suppose that $a,$ $b,$ and $c$ are positive real numbers such that $a^{\log_3 7} = 27,$ $b^{\log_7 11} = 49,$ and $c^{\log_{11} 25} = \sqrt{11}.$ Find $a^{(\log_3 7)^2} + b^{(\log_7 11)^2} + c^{(\log_{11} 25)^2}.$

Respuesta

Solución:

Por la regla de la potencia de una potencia, $\begin{aligned} a^{(\log_3 7)^2} &= \left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7} \\ &= \left(3^{\log_3 7}\right)^3 = 7^3 = 343. \end{aligned}$

De la misma manera, $\begin{aligned} b^{(\log_7 11)^2} &= 49^{\log_7 11} = \left(7^{\log_7 11}\right)^2 \\ &= 11^2 = 121, \end{aligned}$ y $\begin{aligned} c^{(\log_{11} 25)^2} &= \left(\sqrt{11}\right)^{\log_{11} 25} \\ &= \left(11^{\log_{11} 25}\right)^{1/2} \\ &= 25^{1/2} = 5. \end{aligned}$

La suma es $343 + 121 + 5 = 469.$

By the power rule for exponents, $\begin{aligned} a^{(\log_3 7)^2} &= \left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7} \\ &= \left(3^{\log_3 7}\right)^3 = 7^3 = 343. \end{aligned}$

In the same way, $\begin{aligned} b^{(\log_7 11)^2} &= 49^{\log_7 11} = \left(7^{\log_7 11}\right)^2 \\ &= 11^2 = 121, \end{aligned}$ and $\begin{aligned} c^{(\log_{11} 25)^2} &= \left(\sqrt{11}\right)^{\log_{11} 25} \\ &= \left(11^{\log_{11} 25}\right)^{1/2} \\ &= 25^{1/2} = 5. \end{aligned}$

The sum is $343 + 121 + 5 = 469.$

2009 AIME II Problema 2

Solución:

El Problema 2 en otros años