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Mathematics

Assertion (A): log100(1000) = 32\dfrac{3}{2}.

Reason (R): logb a = logxalogxb\dfrac{log_{x}a}{log{x}b}, for all a, b, x.

  1. A is true, R is false.
  2. A is false, R is true.
  3. Both A and R are true.
  4. Both A and R are false.

Logarithms

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Answer

Both A and R are true.

Explanation

Given,

log100(1000) = 32\dfrac{3}{2}

Using logba = logxalogxb\dfrac{logxa}{logxb},

log100(1000) = logx1000logx100\dfrac{logx1000}{logx100}

=logx103logx102=3logx102logx10=3logx102logx10=32= \dfrac{logx10^3}{logx10^2}\\[1em] = \dfrac{3logx10}{2logx10}\\[1em] = \dfrac{3\cancel{logx10}}{2\cancel{logx10}}\\[1em] = \dfrac{3}{2}

Assertion (A) is true.

According to change of base formula of logarithm,

logb a = logxalogxb\dfrac{log{x}a}{log{x}b}

Reason (R) is true.

Hence, both Assertion (A) and Reason (R) are true.

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