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The statement log⁡ax=b\log_a x=bloga​x=b is equivalent to [...]\text{[...]}[...].

A

The product law is log⁡ax+log⁡ay=\log_a x+\log_a y=loga​x+loga​y= log⁡a(xy)\boxed{\log_a(xy)}loga​(xy)​.

B

log⁡a(x2zy)\boldsymbol{\log_a\left(\frac{x^2z}{\sqrt y}\right)}loga​(y​x2z​)

C

The statement log⁡ax=b\log_a x=bloga​x=b is equivalent to ab=x\boxed{a^b=x}ab=x​.

D

A reciprocal: −log⁡ax=log⁡a(1x)-\log_a x=\log_a\left(\frac1x\right)−loga​x=loga​(x1​).

Card 1 of 22

1.6.6 Exponentials and logarithms Flashcards

  1. A Level
  2. /Maths
  3. /1.6.6 Exponentials and logarithms

22 flashcards on WJEC A Level Maths 1.6.6 Exponentials and logarithms: the key formulae, methods and definitions you need to recall for AS Unit 1, AS Unit 2, A2 Unit 3 and A2 Unit 4.

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