Question

Solve `log_2 x − 3 log_(1/2) x` = 6

Answer 1

By change of base rule `log_"b""a" = 1/log_"a""b"`

Given `log_2 x − 3 log_(1/2) x` = 6

`1/log_x 2 - 3/(log_x (1/2))` = 6

`1/log_x 2 - 3/((log_x 1 - log_x 2))` = 6

`1/log_x 2 - 3/(0 - log_x 2)` = 6

`1/log_x 2 + 3/log_x 2` = 6

`4/log_x 2` = 6

log_x 2 = `4/6 = 2/3`

2 = `x^(2/3)`

Raising both sides to the power `3/2` we get

`2^(3/2) = (x^(2/3))^(3/2)`

`2^(3/2) = x^(2/3 xx 3/2)`

⇒ x = `2^(3/2)`

⇒ x = `2^(1 + 1/2)`

=  `2^1*2^(1/2)`

x = `2sqrt(2)`