The number of ordered pairs (x, y) of positiv | vedclass

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Question

(a)

$\because 2^x+3^y=5^{x y}$

When $x=y=1$, then $2+3=5$ and $\left(\frac{2}{5}\right)^x \cdot \frac{1}{5^y}+\left(\frac{3}{5}\right)^y \cdot \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(a)

$\because 2^x+3^y=5^{x y}$

When $x=y=1$, then $2+3=5$ and $\left(\frac{2}{5}ight)^x \cdot \frac{1}{5^y}+\left(\frac{3}{5}ight)^y \cdot \frac{1}{5^x}=1$ here for any $x, y \in Z^{+}$LHS is not equal to 1 as both are less than $\frac{1}{2}$.

$	herefore$ Only one ordered pair $(1,1)$ is possible.

The number of ordered pairs $(x, y)$ of positive integers satisfying $2^x+3^y=5^{x y}$ is

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