Question

if a power b=b power a, show that (a/b)power a/b=a power a/b-1​

Answer 1

Required Answer:-

Given Information:

$\sf \mapsto {a}^{b} = {b}^{a}$

To prove:

$\sf \mapsto { \bigg( \dfrac{a}{b} \bigg)}^{ \frac{a}{b} } = {a}^{ {}^{b}/_{a} - 1 }$

Proof:

Taking LHS,

$\sf { \bigg(\dfrac{a}{b} \bigg)}^{^{a}/_{b} }$

$\sf = \dfrac{ {a}^{{}^{a}/ _{b}}}{ {b}^{^{a}/_{b}} }$

$\sf = \dfrac{ {a}^{{}^{a}/ _{b}}}{ ({b}^{a})^{^{1}/_{b} } }$

We know that,

$\sf \mapsto {a}^{b} = {b}^{a}$

So,

$\sf \dfrac{ {a}^{{}^{a}/ _{b}}}{ ({b}^{a})^{^{1}/_{b} } }$

$\sf = \dfrac{ {a}^{{}^{a}/ _{b}}}{ ( {a}^{b} )^{^{1}/_{b} } }$

$\sf = \dfrac{ {a}^{{}^{a}/ _{b}}}{ ( {a}^{b \times \frac{1}{b}} )}$

$\sf = \dfrac{ {a}^{^{a}/_{b} } }{ {a}^{1} }$

$\sf = {a}^{^{a}/_{b} - 1 }$

Taking RHS,

$\sf = {a}^{^{a}/_{b} - 1 }$

Thus, LHS = RHS (Hence Proved)

Formulae for solving Indices Questions:

$\sf \mapsto({x}^{a} )^{b} = {x}^{ab}$

$\sf \mapsto \dfrac{ {x}^{a} }{ {x}^{b} } = {x}^{a - b} ,x \neq0$

$\sf \mapsto {x}^{a} \times {y}^{a} = {(xy)}^{a}$

$\sf \mapsto \bigg(\dfrac{x}{y} \bigg)^{a} = \dfrac{ {x}^{a} }{ {y}^{a} } ,y \neq0$

$\sf \mapsto {x}^{a} = {x}^{b} \implies a = b,x \neq0$

$\sf \mapsto {x}^{a} = {y}^{a} \implies x = y,a \neq 0$

$\sf \mapsto {x}^{a} = \dfrac{1}{ {x}^{ - a} }$

$\sf \mapsto \sqrt[y]{x} = {x}^{ \frac{1}{y}}$

Answer 2

Answer:

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