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Question : 29 of 86

Marks: +1, -0

If

a^{x-3} \cdot a^{y+2}=a^{2} \cdot a^{x}

a^{x} \cdot a^{y}=a^{4}

then

$x=y=0$
$x=y=1$
$x>y=0$
$x < y$

Solution:

Concept:

Use the law of indices:

a^m \cdot a^n = a^{m+n}

, then equate exponents when bases are equal.

Explanation:

From the first equation:

a^{x-3} \cdot a^{y+2} = a^{2} \cdot a^{x}

.

Left side:

a^{(x-3)+(y+2)} = a^{x+y-1}

.

Right side:

a^{2+x}

.

So

a^{x+y-1} = a^{x+2}

.

Equating exponents:

x+y-1 = x+2

⇒

y=3

.

From the second equation:

a^{x} \cdot a^{y} = a^{4}

⇒

a^{x+y} = a^{4}

⇒

x+y=4

.

Substitute

y=3

:

x+3=4

⇒

x=1

.

Thus

x=1

,

y=3

, so

x < y

.

Answer:

x=1

,

y=3

, so

x < y

. Option D.

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