Asked by Kenny

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.

log_b (x), which means "the logarithm of x, base b."
Answered by Reiny
11/[log_2 (x)] + 1/[2 • log_25 (x)] - 3/[log_8 (x)] = 1/[log_b (x)]

11/log<sub>2</sub>x + 1/[2log<sub>25</sub>x - 3/log<sub>8</sub>x = 1/log<sub>b</sub>x

recall that log<sub>c</sub>d = 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
Answered by oobleck
good answer. It might have been shorter just to note that
log<sub><sub>b</sub></sub>x = log<sub><sub>x</sub></sub>b

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