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Question : 15 of 86

Marks: +1, -0

If

\left(\frac{x^{a}}{x^{b}}\right)^{(a+b)}\left(\frac{x^{b}}{x^{c}}\right)^{(b+c)}\left(\frac{x^{c}}{x^{a}}\right)^{(c+a)}=x

then the value of

x

is

0
$x^{xbc}$
$x^{a+b+c}$
1

Solution:

Concept:

Use the law of exponents:

\frac{x^a}{x^b} = x^{a-b}

and

(x^m)^n = x^{mn}

.

Explanation:

Simplify each factor:

\left(\frac{x^{a}}{x^{b}}\right)^{(a+b)} = \left(x^{a-b}\right)^{a+b} = x^{(a-b)(a+b)} = x^{a^2 - b^2}

.

Similarly,

\left(\frac{x^{b}}{x^{c}}\right)^{(b+c)} = x^{b^2 - c^2}

and

\left(\frac{x^{c}}{x^{a}}\right)^{(c+a)} = x^{c^2 - a^2}

.

Multiplying, add exponents:

(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2) = 0

.

Thus left side equals

x^0 = 1

.

Given that this equals

x

, we have

1 = x

, so

x = 1

.

Answer:

x = 1

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