How do you solve #log_7(x-3)-log_7x=3#?

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Answer 1 · 2016-12-01T05:58:26

#x=-1/114#

Condense the logs on the left -- Subtracting logs of the same base can be rewritten as dividing within the log:
#log_ab-log_ac=log_a(b/c)#

#log_7(x-3)-log_7x=3#
#log_7((x-3)/x)=3#

Now use the log rule: If #log_ab=n#, then #a^n=b#.
#7^3=(x-3)/x#

#343=(x-3)/x#

Multiply each side by x:
#343x=x-3#

Subtract x from each side:
#342x=-3#

#x=-3/342#

#x=-1/114#