Asked by Ann
If log base a * b = 1/x
and log base b * sq.rt a = 3x^2 show that x=1/6.
and log base b * sq.rt a = 3x^2 show that x=1/6.
Answers
Answered by
Reiny
by definition
log<sub>a</sub>b = log b/log a
then log<sub>b</sub>√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
log<sub>a</sub>b = log b/log a
then log<sub>b</sub>√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
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