Note that the subscript is b and not 6. There is no subscript available for b for Unicode.

1 / logₐ₆ b - 1 / logₐ₆a

= log₆ab - logₐab

= log₆a + log₆b - (logₐa + logₐb)

= log₆a + 1 - 1 - logₐb

= log₆a - logₐb



Given :

1 / logₐb + 1 / log₆a = √293

log₆a + logₐb = √293


Square both sides,

(log₆a + logₐb)² = 293

(log₆a)² + 2 (log₆a) (logₐb) + (logₐb)² = 293

(log₆a)² + 2 (log₆a) (1 / log₆a) + (logₐb)² = 293

(log₆a)² + 2 + (logₐb)² = 293

(log₆a)² + (logₐb)² + 2 - 4 = 293 - 4

(log₆a)² - 2 - (logₐb)² = 289

(log₆a)² - 2 (log₆a) (logₐb) + (logₐb)² = 289

(log₆a - logₐb)² = 289

log₆a - logₐb = √289 or -√289

log₆a - logₐb = 17 or -17

[ Now -17 is rejected.

Since a > b > 1 , both log₆a and logₐb are positive (>0) , but log₆a > 1 and 0 < logₐb < 1

So log₆a > logₐb
Which in turn means log₆a - logₐb > 0 , so it cannot be negative)


Eg. If a = 6 and b = 5 ,

Then log₅6 ≈ 1.113 but log₆5 ≈ 0.898 . So log₅6 - log₆5 ≈ 0.215 > 0 ]


So 1 / logₐ₆ b - 1 / logₐ₆a = log₆a - logₐb = 17