The value of 0-05log-20 0-1 - 0-01 - 0-001 -

The correct option is A 81(0.05)^log_ √(20) (0.1 + 0.01 +.....) = (1/20)^2 log_20 (0.1/1 0.1) = 20^ 2 log_20 (1/9) = 20^2 log_20 9 = 20^log_20 9^2 = 9^2 ...

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Byju's Answer

Standard X

Mathematics

Fundamental laws of logarithms

The value of ...

Question

The value of $(0.05)_{\sqrt{20}}^{l o g} (0.1 + 0.01 + 0.001 + . . . . .)$

81

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\frac{1}{81}

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20

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\frac{1}{20}

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Solution

The correct option is A $81$
$(0.05)_{\sqrt{20}}^{l o g} (0.1 + 0.01 + . . . . .) = (\frac{1}{20})^{2 l o g_{20} (\frac{0.1}{1 - 0.1})} = 20^{- 2 l o g_{20} (\frac{1}{9})} = 20^{2 l o g_{20} 9} = 20^{l o g_{20} 9^{2}} = 9^{2} = 81.$

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{(0.05)}^{{log}_{\sqrt{20}} (0.1 + 0.01 + 0.001 + . . . .)}

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The value of (0.05)log√20 (0.1+0.01+0.001+.....)

The correct option is A 81(0.05)log√20 (0.1+0.01+.....)=(120)2log20 (0.11−0.1)=20−2log20 (19)=202 log20 9=20log20 92=92=81.

The value of $(0.05)_{\sqrt{20}}^{l o g} (0.1 + 0.01 + 0.001 + . . . . .)$

The correct option is A $81$
$(0.05)_{\sqrt{20}}^{l o g} (0.1 + 0.01 + . . . . .) = (\frac{1}{20})^{2 l o g_{20} (\frac{0.1}{1 - 0.1})} = 20^{- 2 l o g_{20} (\frac{1}{9})} = 20^{2 l o g_{20} 9} = 20^{l o g_{20} 9^{2}} = 9^{2} = 81.$