Math, asked by aliahmad46, 6 months ago

please answer only if u know... please dont post irrelevant answers​

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Answered by bson
1

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

Answered by sameeha343
16

Answer:

Thanks for answering my question

thank you so much dear

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