Question

Can you please help me with this

Answer 1

Answer:

log9

Step-by-step explanation:

Using the rules of logarithms

logx + logy = log(xy)

logx - logy = log($\frac{x}{y}$)

log$x^{n}$ ⇔ nlogx

Given

2(log18- log3) + $\frac{1}{2}$log$\frac{1}{16}$

= 2(log($\frac{18}{3}$ ) ) + log$(\frac{1}{16}) ^{\frac{1}{2} }$

= 2log6 + log$\frac{1}{4}$

= log6² + log$\frac{1}{4}$

= log36 + log$\frac{1}{4}$

= log( 36 × $\frac{1}{4}$ )

= log9