Question

Please answer this question as soon as possible
If a^b=c^d ; prove that [1/a]^b/d * [c/a]^[d+b/b]=1

Answer 1

Solution :

$\displaystyle \sf{ \bold { To \: Prove \: - }}$

$\displaystyle \sf{ \implies{ \bold { ( \dfrac{1}{a} )^{ (\dfrac{b}{d} ) } \times \dfrac{ c }{ a }^{ ( \dfrac{ d + b }{ b} ) } = 1}}}$

$\\ \\ \sf{ \orange { Proof \: - }} \\ \\ \monospace { \bold { We \: know \: that \: - }} \\ \\ \tt{ \blue { 1^{(any \: number )} = 1 \: only }} \\ \\ \tt{ \red { \implies { a^{(-b/d )} \times \dfrac{ c }{ a }^{ ( \dfrac{ d + b }{ b} ) } }}} \\ \\ \tt{ \green { a^b = c^d }} \\ \\ \tt{ \orange { a = c^{(b/d)} }} \\ \\ \tt{ \blue { \implies { c^{(-b^2/d^2 )} \times c^{ ( d/b + 1 - b/d ) } }}} \\ \\ \tt{ \red { \implies { c^0 = 1 }}} \\ \\ \tt{ \monospace { Hence \: Proved }}$

Answer 2

Required Proof :-

$\small \sf{ \dag \: As \: we \: know \: that}$

$\sf \: {1}^{(any \: number)} = 1$

For example =>

1² = 1 × 1 = 1

1³ = 1 × 1 × 1 = 1

$\sf \: a {}^{ \frac{ - b}{d} } \times \dfrac{c}{a} {}^{ \bigg(\dfrac{d + b}{b} \bigg)}$

$\sf \: a = {c}^{ \frac{b}{d} }$

$\sf \: {c}^{ \dfrac{ - {b}^{2} }{ {d}^{2} } } \times c {}^ {\dfrac{d}{b} +1 - \dfrac{b}{d} }$

$\sf \: {c}^{0} = 1$