Question #0e56f

2 Answers
Jim S
Nov 20, 2017

#*# #x=0# or #x=3#

Explanation:

#log(3x+1)-log(x^2+1)=##log((3x+1)/(x^2+1))#

#*# #log((3x+1)/(x^2+1))=0# #<=># #log((3x+1)/(x^2+1))=log(1)#

log is #1-1# so

#(3x+1)/(x^2+1)# #=1# #<=># #3x+1=x^2+1# #<=># #x^2-3x=0# #<=>#

#x(x-3)=0#

#*# #x=0# or #x=3#

Jim G.
Nov 20, 2017

#x=0" or "x=3#

Explanation:

#"using the "color(blue)"law of logarithms"#

#•color(white)(x)logx-logy=log(x/y)#

#rArrlog((3x+1)/(x^2+1))=0#

#"note that "log" usually denotes "log" to base 10"#

#rArr((3x+1)/(x^2+1))=10^0=1#

#rArrx^2+1=3x+1#

#rArrx^2-3x=0#

#rArrx(x-3)=0#

#rArrx=0" or "x=3#

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