Question

a) log5cos0*

b) log2(cos120*+2sin150*)

c) log8sin120* − log8tg60*

Answer 1

a) log5cos0*=log₅1= 0

5^x=1

5^x=5^0

x=0

 

b) log2(cos120*+2sin150*)=log₂(-½+2*½)=log₂½= -1

2^x=½

2^x=2⁻¹

x=-1

 

c) log8sin120* − log8tg60*=log₈√3/2 - log₈√3=-⅓