How do you solve #\log _ { 5} ( x + 1) - \log _ { \frac { 1} { 5} } ( 2x + 5) < \log _ { 5} ( 2x ^ { 2} - x + 9)#?

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Answer 1 · 2018-06-23T13:41:21

#x<1/2#

Writing

#ln(x+1)-ln(2x+5)/ln(1/5)< ln(2x^2-x+9)/ln(5)#

we have used that #log_ab=ln(b)/ln(a)#
and #ln(1/5)=ln(1)-ln(5)=-ln(5)#
so we get

#ln((x+1)(2x+5)) < ln(2x^2-x+9)# since #ln(5)>0#

so we get

#2x^2+2x+5x+5 < 2x^2-x+9#

simplifying

#8x<4#

#x<1/2#