How do you solve #5log_10(x-2)=11#?

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Answer 1 · 2018-07-03T13:20:58

#color(blue)(x=10^(11/5)+2~~160.4893192)#

#5log_(10)(x-2)=11#

Divide by 5:

#log_(10)(x-2)=11/5#

Raising the base to these:

#10^(log_(10)(x-2))=10^(11/5)#

#x-2=10^(11/5#

#x=10^(11/5)+2~~160.4893192#