Find the sum of all value(s) of b such that 11/[log_2 (x)] + 1/[2 • log_25 (x)] - 3/[log_8 (x)] = 1/[log_b (x)] for all x > 1.

Question

log_b (x), which means "the logarithm of x, base b."

Answer 1

11/[log_2 (x)] + 1/[2 • log_25 (x)] - 3/[log_8 (x)] = 1/[log_b (x)]

11/log₂x + 1/[2log₂₅x - 3/log₈x = 1/log_bx

recall that log_cd = log d/log e , base 10 for both
so our equation becomes
11log2/logx + (1/2)log25/logx - 3log8/logx = logb/logx
multiply every term by logx
11log2 + (1/2)log25 - 3log8 = logb
log(2^11) + log(25^(1/2)) - log8^3 = logb/logx
log(2^11 * 5 / 8^3) = logb
log 20 = log b
b = 20

check my arithmetic

Answer 2

good answer. It might have been shorter just to note that
log_bx = log_xb

Saves going through the change of base. But that's always a good step to know, anyway.