Question

find the multiplicative inverse of (7/8)^-3 ÷(8/7)^5​

Answer 1

$\huge{\bf{\red{Solution}}}$

[ ( 7 / 8 )^-3 ] / [ ( 8 / 7 ) ^5 ]

= [ ( 7 / 8 ) ^-3 ] X [ ( 7 / 8 ) ^5 ]

When the bases are equal the powers should be added

= [ ( 7 / 8) ^ 2 ]

= 49 / 64

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