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Exercise 2.12 · Q2

Q.Compute log⁡927−log⁡279\log_9 27-\log_{27}9.

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✓ Free question

Step 1. log⁡927=log⁡3233=32\log_9 27=\log_{3^2}3^3=\dfrac32 (using log⁡aman=nm\log_{a^m}a^n=\dfrac nm).

Step 2. log⁡279=log⁡3332=23\log_{27}9=\log_{3^3}3^2=\dfrac23.

Step 3. log⁡927−log⁡279=32−23=96−46=56\log_9 27-\log_{27}9=\dfrac32-\dfrac23=\dfrac96-\dfrac46=\dfrac56.

✓Final answer

56\dfrac56.

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