How do you solve # log x- log(x-8)=1#?

1 Answer

#x=80/9=8 8/9#

Explanation:

The given:# log x- log (x-8) = 1#

This means the common logarithm of base 10:

#log_10 x-log_10(x-8) = 1#

Also #log_10 10=1#

therefore

#log_10 x-log_10(x-8) = log_10 10#

#log_10 (x/(x-8))=log_10 10#

Take Antilogarithm of both sides of the equation

#Antilog(log_10 (x/(x-8)))=Antilog(log_10 10) #

means

#10^(log_10 (x/(x-8)))=10^(log_10 10)#

means

#x/(x-8)=10#

solve for #x# now

#x=10(x-8)#

#x=10x-80#

#x-10x=-80#

#-9x=-80#

#(-9x)/-9=(-80)/-9#

#x=80/9=8 8/9#

Have a nice day !!! from the Philippines..