if (3/8)^-5×(16/21)^-5=(2/7)^x find the x^3
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Answered by
67
(3/8)^-5 × ( 16/21)^-5 = (2/7)^x
[ use, x^m.y^n = (xy)^mn]
{3/8 × 16/21}^-5 = (2/7)^x
(2/7)^-5 = (2/7)^x
[ use, P^m = P^n then, m = n ]
-5 = x
hence, x³ = (-5)³ = -125
[ use, x^m.y^n = (xy)^mn]
{3/8 × 16/21}^-5 = (2/7)^x
(2/7)^-5 = (2/7)^x
[ use, P^m = P^n then, m = n ]
-5 = x
hence, x³ = (-5)³ = -125
Answered by
36
(3/8)^-5×(16/21)^-5=(2/7)^x
[ (a^m × b^m) =(ab)^m
(3/8× 16/21)^-5 = (2/7)^x
(2/7)^-5 = (2/7)^x
[a^m = a^n then m=n]
x= -5
x³= (-5)³
x= -125
[ (a^m × b^m) =(ab)^m
(3/8× 16/21)^-5 = (2/7)^x
(2/7)^-5 = (2/7)^x
[a^m = a^n then m=n]
x= -5
x³= (-5)³
x= -125
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