How do you solve these logarithms?

Question

#2 log_(2)(1-2x)=12#
#log(7x)+4=5#

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Answer 1 · 2016-11-26T22:44:12

#x=(1pm2^6)/2#
#x=e/7#

Before the #log# application we had:

#(1-2x)^2=2^12#
#7xe^4=e^5#

so

#1-2x=pm2^6->x=(1pm2^6)/2#

and

#x=e/7#