Question

[(-3by4)^5]^0 = (-3by4)^2x[(−3by4)5]0=(−3by4)2x
$\ \textless \ br /\ \textgreater \ \ \textless \ br /\ \textgreater \ [(-3by4)^5]^0 = (-3by4)^2x[(−3by4)5]0=(−3by4)2x\ \textless \ br /\ \textgreater$
Do please .........
Give Right answer
serial wise ok​

Answer 1

Answer:

x = 0

Step-by-step explanation:

[(-3/4)^5]^0 = (-3/4)^2x

1 = (-3/4)^2x

2x = 0

x = 0

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Answer 2

Answer:

x=0

Step-by-step explanation:

$[(-\frac{3}{4} )^5]^0=(-\frac{3}{4})^{2x}$

=>$(-\frac{3}{4}) ^{5*0}=(-\frac{3}{4} )^{2x}$

=>$(-\frac{3}{4} )^0=(-\frac{3}{4} )^{2x}$

comparing both , we get

2x=0

=>x=0

HOPE IT HELPS,

PLEASE THANK , FOLLOW AND MARK AS BRAINLIEST.