证明：∵log₁₂27=a，
∴

4(3−a)

3+a

=

4(3−3log123)

3+3log123

=

4(1−log123)

1+log123

=

log12124−log1234

log12 36

=

log12256

log1236

=log₃₆256
=log₆16．
∴log₆16=

4(3−a)

3+a

．