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Answers
Step-by-step explanation:
let log a = m; log b = n base=10
log b = log b / log a assume base =10
a
= n/m ----A
because log m base n = logm / log n
log b^½ base a^½/b³ =log b^½ /( log a^½ - log b³)
=½log b / (½Log a -3 log b)
= n/2/(m/2 -3n) ---- B
because log(a/b)=log a- log b and log mⁿ= n log m
log b base a/b⁴ = log b / log(a/b⁴)
= log b / (log a - 4log b)
= n/(m-4n) ---C
log b base a/b⁶ = log b / log a/b⁶
= log b/ ( log a - 6 log b)
= n/ (m-6n) ----D
log a³ b^-12 base b= log a³+ log b^-12 / log b
= 3 log a -12 log b / log b
=(3m -12n) / n
= 3(m-4n)/n ---F
A-B = n/m - n/2/ (m/2 - 3n)
= n/m - n/(m-6n) = n(1/m - 1/(m-6n))
= -6n²/ m(m-6n)
C-D = n / (m-4n) - n / (m-6n) = n(1/(m-4n) - 1/(m-6n))
= n(-2n/ (m-4n)(m-6n)) = -2n²/(m-4n)(m-6n)
A-B / C-D = -6n²/ m(m-6n) ÷ -2n²/(m-4n)(m-6n)
=3x(m-4n)(m-6n)/m(m-6n)
=3 × (m-4n)/m ---E
E/F = 3(m-4n)/m ÷ 3(m-4n)/n
=n/m
=log b/ log a
=log b
a
Answer:
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