How do you solve #log (x+4)=log x + log 4#?

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Answer 1 · 2015-08-15T15:40:49

#log(x+4) = log(x) + log(4) = log(4x)#, so #x+4 = 4x#, hence #x = 4/3#

If #a, b > 0# then #log(a)# and #log(b)# are defined and

#log(a) + log(b) = log(ab)#

So #log(x) + log(4) = log(4x)#

So our equation becomes:

#log(x + 4) = log x + log 4 = log(4x)#

Since #log:(0, oo) -> RR# is one-one, that means:

#x+4 = 4x#

Subtract #x# from both sides to get #4 = 3x#, then divide both sides by #3# to get #x = 4/3#.