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Maths-

General

Easy

Question

If $open parentheses 21.4 close parentheses to the power of a end exponent equals open parentheses 0.00214 close parentheses to the power of b end exponent equals 100 comma$ then the value of $fraction numerator 1 over denominator b end fraction minus fraction numerator 1 over denominator b end fraction$ is

0
1
2
4

The correct answer is: 2

$open parentheses 21.4 close parentheses to the power of a end exponent equals 100 rightwards double arrow a log invisible function application open parentheses 21.4 close parentheses equals 2$
$therefore log invisible function application open parentheses 21.4 close parentheses equals 2 divided by a blank open parentheses i close parentheses$
Again $open parentheses 0.00214 close parentheses to the power of b end exponent equals 100 comma blank w e blank g e t blank b open parentheses log invisible function application 0.00214 close parentheses equals 2$
$rightwards double arrow b log invisible function application open parentheses 21.4 cross times 10 to the power of negative 4 end exponent close parentheses equals 2$
$rightwards double arrow b equals fraction numerator 2 over denominator log invisible function application 21.4 minus 4 end fraction equals fraction numerator 2 over denominator fraction numerator 2 over denominator a end fraction minus 4 end fraction equals fraction numerator 1 over denominator fraction numerator 1 over denominator a end fraction minus 2 end fraction$
$rightwards double arrow b equals fraction numerator a over denominator 1 minus 2 a end fraction semicolon fraction numerator 1 over denominator b end fraction equals fraction numerator 1 minus 2 a over denominator a end fraction$
$rightwards double arrow fraction numerator 1 over denominator a end fraction minus fraction numerator 1 over denominator b end fraction equals 2$

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