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Exponents Of Real Numbers — Maths STD 9 — Question

Gujarat BoardEnglish MediumSTD 9MathsExponents Of Real Numbers1 Mark
Question
Prove that:
$\Big(\frac{\text{x}^\text{a}}{\text{x}^\text{b}}\Big)^\text{c}\times\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{c}}\Big)^\text{a}\times\Big(\frac{\text{x}^\text{c}}{\text{x}^\text{a}}\Big)^\text{b}=1$
✓

Answer

To prove,
$\Big(\frac{\text{x}^\text{a}}{\text{x}^\text{b}}\Big)^\text{c}\times\Big(\frac{\text{x}^\text{b}}{\text{x}^\text{c}}\Big)^\text{a}\times\Big(\frac{\text{x}^\text{c}}{\text{x}^\text{a}}\Big)^\text{b}=1$
Left hand side $(LHS) =$ Right hand side $(RHS)$ Considering $LHS,$
$=\Big(\frac{\text{x}^{\text{ac}}}{\text{x}^{\text{bc}}}\Big)\times\Big(\frac{\text{x}^{\text{ba}}}{\text{x}^{\text{ca}}}\Big)\times\Big(\frac{\text{x}^{\text{bc}}}{\text{x}^{\text{ab}}}\Big)$
$=\text{x}^{\text{ac}-\text{bc}}\times\text{x}^{\text{ba}-\text{ca}}\times\text{x}^{\text{bc}-\text{ab}}$
$=\text{x}^{\text{ac}-\text{bc}+\text{ba}-\text{ca}+\text{bc}-\text{ab}}$
$=\text{x}^0$
$=1$
Therefore, $LHS = RHS$
Hence proved.

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