If 34x-81-1 and 101-y - 0-0001- find the value of 2-x- 16y

3^4x=(81)^ 1 (3^4)^x=(81)^ 1 (81)^x=(81)^ 1 x = 1 (10)^1/y = 0.0001 = 10^ 4 1/y = 4 y = 1/4 2^ x× 16^y = 2^ ( 1)× 16^ 1/4 = 2 ×1/16^1/4 = 2 ×1/2 = ...

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Byju's Answer

Standard IX

Mathematics

Power Law

If 34x=81-1 a...

Question

If $3^{4 x} = (81)^{- 1}$ and $(10)^{\frac{1}{y}} = 0.0001$ , find the value of $2^{- x} \times 16^{y}$

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Solution

$3^{4 x} = (81)^{- 1} (3^{4})^{x} = (81)^{- 1} (81)^{x} = (81)^{- 1} x = - 1$

$(10)^{\frac{1}{y}} = 0.0001 = 10^{- 4} \frac{1}{y} = - 4 y = - \frac{1}{4}$

$2^{- x} \times 16^{y} = 2^{- (- 1)} \times 16^{- \frac{1}{4}} = 2 \times \frac{1}{16^{\frac{1}{4}}} = 2 \times \frac{1}{2} = 1$

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If 34x=(81)−1 and (10)1y=0.0001, find the value of 2−x×16y

34x=(81)−1(34)x=(81)−1(81)x=(81)−1x=−1 (10)1y=0.0001=10−41y=−4y=−14 2−x×16y=2−(−1)×16−14=2×11614=2×12=1

If $3^{4 x} = (81)^{- 1}$ and $(10)^{\frac{1}{y}} = 0.0001$ , find the value of $2^{- x} \times 16^{y}$

$3^{4 x} = (81)^{- 1} (3^{4})^{x} = (81)^{- 1} (81)^{x} = (81)^{- 1} x = - 1$

$(10)^{\frac{1}{y}} = 0.0001 = 10^{- 4} \frac{1}{y} = - 4 y = - \frac{1}{4}$

$2^{- x} \times 16^{y} = 2^{- (- 1)} \times 16^{- \frac{1}{4}} = 2 \times \frac{1}{16^{\frac{1}{4}}} = 2 \times \frac{1}{2} = 1$