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

Escribe

a^{(\log_3 7)^2}

como

\left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7}

Write

a^{(\log_3 7)^2}

as

\left(a^{\log_3 7}\right)^{\log_3 7} = 27^{\log_3 7}

27^{\log_3 7} = \left(3^{\log_3 7}\right)^3 = 7^3.

Trata los otros dos términos de la misma manera.

27^{\log_3 7} = \left(3^{\log_3 7}\right)^3 = 7^3.

Handle the other two terms the same way.

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)^{\frac{1}{2}} \\ &= 25^{\frac{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)^{\frac{1}{2}} \\ &= 25^{\frac{1}{2}} = 5. \end{aligned}

The sum is

343 + 121 + 5 = 469.