How do you solve #10^(x+2) - 12 = 22 # and find any extraneous solutions?

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Answer 1 · 2016-07-22T10:18:07

The Soln. is # x=-0.4685#

#10^(x+2)-12=22 rArr 10^(x+2)=12+22=34#

#rArrlog_(10)10^(x+2)=log_(10)34#

#rArr (x+2)log_(10)10=1.5315#..................[Using LogTable]

#rArr (x+2)*1=1.5315#

#rArr x+2=1.5315#

#rArr x=1.5315-2#

#rArr x=-0.4685# is the Soln.