How do you solve these logarithmic equations?

#4+log_(9)(3x-7)=6#
#log_(2)(2x)+log_(2)x=5#
#3log_(5)x-log_(5)(5x)=3-log_(5)25#

2 Answers
Cesareo R.
Nov 25, 2016

See below.

Explanation:

Before the logarithm application, the equations read

#9^4(3x-7)=9^6#
#(2x)x=2^5#
#x^3/(5x)=5^3/25#

after that we have

#3x-7=9^2->x=(9^2+7)/3#
#x^2=2^4->x=pm2^2#
#x^2=5^2->x=pm 5#

Monzur R.
Nov 25, 2016

#4+log_9(3x-7)=6#

#log_9(3x-7)=2#

#3x-7=81#

#3x=89#

#x=89/3#

#log_2(2x)+log_2(x)=5#

#log_2(2x^2)=5#

#2x^2=2^5=32#

#x^2=16#

#x=+-4#

#3log_(5)x-log_(5)(5x)=3-log_(5)25#

#log_5x^3-log_5(5x)=3-2=1#

#log_5(x^3/(5x))=log_5(x^2/5)=1#

#x^2/5=5#

#x^2=25#

#x=+-5#

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