#81^(1/(log _5 3)) +27^((log _9 36)) + 3^(1/(log _7 9))#?

Question

26631 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-08-29T07:38:33

#81^(1/(log_5 3)) + 27^(log_9 36) + 3^(1/(log_7 9)) = 841+sqrt(7)#

#1/(log_5 3) = log_3(5)#

So:

#81^(1/(log_5 3)) = (3^4)^(log_3(5)) = 3^(4log_3(5)) = (3^(log_3(5)))^4 = 5^4 = 625#

#log_9 36 = (log_3(36))/(log_3(9)) = log_3(36)/(log_3(3^2)) = log_3(36)/2#

So:

#27^(log_9 36) = (3^3)^(log_3(36)/2) = (3^(log_3(36)))^(3/2) = 36^(3/2) = (6^2)^(3/2) = 6^3 = 216#

#1/(log_7 9) = log_9(7) = log_3(7)/2#

So:

#3^(1/(log_7 9)) = 3^(log_3(7)/2) = (3^(log_3(7)))^(1/2) = 7^(1/2) = sqrt(7)#

Putting this all together:

#81^(1/(log_5 3)) + 27^(log_9 36) + 3^(1/(log_7 9)) = 625 + 216 + sqrt(7) = 841+sqrt(7)#