Question

Supongamos que

4^{x_1} = 5,

5^{x_2} = 6,

6^{x_3} = 7,

\ldots,

127^{x_{124}} = 128.

¿Cuánto vale

x_1 x_2 \cdots x_{124}

?

Suppose that

4^{x_1} = 5,

5^{x_2} = 6,

6^{x_3} = 7,

\ldots,

127^{x_{124}} = 128.

What is

x_1 x_2 \cdots x_{124}?

2

\dfrac{5}{2}

3

\dfrac{7}{2}

4

Cada ecuación da

x_k = \log_{k+3}(k+4)

.

Each equation gives

x_k = \log_{k+3}(k+4)

El producto se reduce telescópicamente a

\log_4 128

.

The product telescopes to

\log_4 128

De

4^{x_1} = 5

obtenemos

x_1 = \log_4 5,

y en general

x_k = \log_{k+3}(k+4).

El producto se reduce telescópicamente:

\begin{aligned} &x_1 x_2 \cdots x_{124} \\ &= \log_4 5 \cdot \log_5 6 \cdots \log_{127} 128 \\ &= \log_4 128. \end{aligned}

Como

128 = 2^7

y

4 = 2^2,

esto es igual a

\dfrac{7\log 2}{2\log 2} = \dfrac72.

Por lo tanto, la respuesta correcta es D.

From

4^{x_1} = 5

we get

x_1 = \log_4 5,

and in general

x_k = \log_{k+3}(k+4).

The product telescopes:

\begin{aligned} &x_1 x_2 \cdots x_{124} \\ &= \log_4 5 \cdot \log_5 6 \cdots \log_{127} 128 \\ &= \log_4 128. \end{aligned}

Since

128 = 2^7

and

4 = 2^2,

this equals

\dfrac{7\log 2}{2\log 2} = \dfrac72.

Thus, the correct answer is D.