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

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Answer 1 · 2016-02-07T16:17:19

Put all logs to one side of the equation and solve using the property #log_an - log_am = log_a(n/m)#

#log(x + 4) - logx = log4#

#log(((x + 4)/(x))/4 )= 0#

Convert to exponential form. The base is 10, since nothing is noted in subscript in the log.

#(((x + 4)/x) / 4) = 10^0#

#(x + 4)/x = 1 xx4#

x + 4 = 4x

4 = 3x

#4/3# = x

Hopefully this helps.

Answer 2 · 2016-02-07T16:49:10

#x=4/3#

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

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

#log((x+4)/x)-log(4)=0#

#log((x+4)/x xx1/4)=0#

#log((x+4)/(4x))=0#

Consider standard form: #log_10(x)=y -> 10^y=x#

so#" " log_10((x+4)/(4x))=0 -> 10^0=(x+4)/(4x)#

Giving: #" "4x=x+4#

#3x=4#

#x=4/3#