Schoolcrop

(Q37). We know that (a)^mn = a^mn let a^m = x then m = log_ax , xⁿ = a^mn then log_axⁿ = mn= nlog_ax (why)

(a)^mn = a^mn a^m = x

log _axⁿ = log a (a^m)ⁿ(a^m)ⁿ = a^mn

= log_aa^mn

= mn log_aa

mn (1) = mn

log_axⁿ = mn .........1

n log_xx = n log_aa^m

= n (m log_aa)

= n (m ()) = mn

log_ax = mn .........2

Here 1 & 2 are

Log_axⁿ = mn = log_ax

Verified

Log_axⁿ = mn = log_ax