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प्रश्न
Solve the following equation for x:
`4^(2x)=1/32`
उत्तर
`4^(2x)=1/32`
`rArr(2^2)^(2x)=1/2^5`
`rArr2^(4x)xx2^5=1`
`rArr2^(4x+5)=2^0`
⇒ 4x + 5 = 0
⇒ 4x = -5
`rArr x=-5/4`
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