Question

Express as a single logarithm:

`2/3 log 8 + 6 log root(3)(2) - 1/2 log  1/9`

Answer 1

`2/3 log 8 + 6 log root(3)(2) - 1/2 log  1/9`

Using the rule n log ⁡a = log ⁡aⁿ,

`log 8^(2/3) + log(root3 2)^6 - log(1/9)^(1/2)`

Simplify the powers:

⇒ `8^(2/3)` 

= `(2^3)^(2/3)`

= 2²

= 4

⇒ `(root(3)(2))^6` 

= `2^(6/3)`

= 2²

= 4

⇒ `(1/9)^(1/2)`

= `1/3`

∴ `log 4 + log 4 - log  1/3`

Using the rules log ⁡a − log ⁡b = `log⁡(a/b)` and log⁡ a + log ⁡b = log ⁡(ab),

= `log ((4 xx 4)/(1/3))`

= log (16 × 3)

= log 48