Find -x- if 8x-2-14-3 x-2-0-0625x-

The correct option is D 1 8^x 2× (1/2)^4 3x = (0.0625)^x 8^x 2×8^x/16 = (1/16)^x 8^2x 2/16 = (1/16)^x For x=1 8^2 × 1 2/16 = 1/16 1/16 = 1/16 ∴ x = ...

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Byju's Answer

Standard IX

Mathematics

Multiplication of Numbers with Square Roots

Find 'x', if ...

Question

Find 'x', if $8^{x - 2} \times {\frac{1}{2}}^{4 - 3 x} = (0.0625)^{x}$ .

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Solution

The correct option is D
1

$8^{x - 2} \times (\frac{1}{2})^{4 - 3 x} = (0.0625)^{x} 8^{x - 2} \times \frac{8^{x}}{16} = (\frac{1}{16})^{x} \frac{8^{2 x - 2}}{16} = (\frac{1}{16})^{x} F o r x = 1 \frac{8^{2 \times 1 - 2}}{16} = \frac{1}{16} \frac{1}{16} = \frac{1}{16}$
$∴$ x = 1

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Find 'x', if 8x−2×124−3x=(0.0625)x.

The correct option is D 1 8x−2×(12)4−3x=(0.0625)x8x−2×8x16=(116)x82x−216=(116)xFor x=182×1−216=116116=116 ∴ x = 1

Find 'x', if $8^{x - 2} \times {\frac{1}{2}}^{4 - 3 x} = (0.0625)^{x}$ .

The correct option is D
1

$8^{x - 2} \times (\frac{1}{2})^{4 - 3 x} = (0.0625)^{x} 8^{x - 2} \times \frac{8^{x}}{16} = (\frac{1}{16})^{x} \frac{8^{2 x - 2}}{16} = (\frac{1}{16})^{x} F o r x = 1 \frac{8^{2 \times 1 - 2}}{16} = \frac{1}{16} \frac{1}{16} = \frac{1}{16}$
$∴$ x = 1