How do you solve #5^(x+3)>10^(x-6)#?

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Answer 1 · 2016-11-08T19:03:08

#x < 6+9log_2 5#

#10^(x-6)=2^(x-6)5^(x-6)# then

#5^(x+3)/(5^(x-6)) > 2^(x-6)# but

#5^(x+3)/(5^(x-6)) =5^(x+3-(x-6)) = 5^9# then

#5^9 > 2^(x-6)# or

#5^9 2^6 > 2^x# now aplying #log# to both sides the inequality is preserved because #log# is monotonic strict increasing. So

#9log_2 5+6log_2 2>x# or finally

#x < 6+9log_2 5#