Show that: log √x with base of 1/√b = -logx with base b

Answer 1

let k=log_√(1/b)√x

then √x=(1/√b)^k=b^(-k/2)

x=b^-k

log_bx=-k

-log_bx=log_√(1/b)√x

Answer 2

Using the change of base formula, log_(1/√b√x = log_b √x / log_b(1 / √b) = (1/2)log_bx / [log_b1 - (1/2)log_bb] =

(1/2)og_bx / (-(1/2)) = -log_bx.