Tolong krjakn dong plizzzzz

a. ⁴log1/9 = ⁴log3⁻² = -2 . ⁴log3 = -2x
b. ²log√3 = [tex] ^{4^{ \frac{1}{2} } }log 3^{ \frac{1}{2} } = \frac{ \frac{1}{2} }{ \frac{1}{2} } .^{4}log3 = ^{4}log3 = [/tex] x