Question

Can you please help me with this

Answer 1

log9

Explanation:

Using the rules of logarithms

logx + logy = log(xy)

logx - logy = log(
`(x)/(y)`)

log
`x^(n)` ⇔ nlogx

Given

2(log18- log3) +
`(1)/(2)`log
`(1)/(16)`

= 2(log(
`(18)/(3)` ) ) + log
`((1)/(16)) ^{(1)/(2) }`

= 2log6 + log
`(1)/(4)`

= log6² + log
`(1)/(4)`

= log36 + log
`(1)/(4)`

= log( 36 ×
`(1)/(4)` )

= log9