Answer:
log9
Step-by-step explanation:
Using the rules of logarithms
logx + logy = log(xy)
logx - logy = log([tex]\frac{x}{y}[/tex])
log[tex]x^{n}[/tex] ⇔ nlogx
Given
2(log18- log3) + [tex]\frac{1}{2}[/tex]log[tex]\frac{1}{16}[/tex]
= 2(log([tex]\frac{18}{3}[/tex] ) ) + log[tex](\frac{1}{16}) ^{\frac{1}{2} }[/tex]
= 2log6 + log[tex]\frac{1}{4}[/tex]
= log6² + log[tex]\frac{1}{4}[/tex]
= log36 + log[tex]\frac{1}{4}[/tex]
= log( 36 × [tex]\frac{1}{4}[/tex] )
= log9
