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Solve for #x#, given #x^((log_5 x)-2) = (x^2/(125))# ?

1 Answer

Oct 23, 2017

#x = {5,125}#

Explanation:

Applying #log_5# to both sides

#(log_5 x-2)log_5 x=2log_5x - 3# and now calling #y = log_5 x#

#(y-2)y=2 y-3# or #y^2-4y+3=0# now solving for #y#

#y = {1,3}# or

#log_5 x = 1 rArr x = 5# and #log_5 x = 3 rArr x = 125#

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