Question

If log base a * b = 1/x

and log base b * sq.rt a = 3x^2 show that x=1/6.

Answer 1

by definition

log_ab = log b/log a

then log_b√a
= log √a/log b
= (1/2)log a/log b

lets form the product
[log b/log a][(1/2)log a/log b) = (1/x)(3x^2)
1/2 = 3x
1 = 6x
x = 1/6