How do you solve #log3x=log2+log (x+5)#?

Question

4974 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-04-08T10:58:51

#x = 10#

Rearrange slightly to get #x# on one side and constant on the other

#log3x = log 2 + log(x + 5)#
#log3 + logx = log 2 + log(x+5)#
#logx - log(x+5) = log2 - log3#

Now, using laws of logarithms, make it so you only have a single logarithm on either side

#logx - log(x+5) = log(x/(x+5))#
#log2 - log3 = log(2/3)#

So you have

#log(x/(x+5)) = log(2/3)#

Raise both sides to the power #e# to cancel out the logarithms (assuming we are dealing with #log_e# or #ln#),

#x/(x+5) = 2/3#

From which we get

#x/(x+5) = 2/3#
#3x = 2(x+5)#
#3x = 2x + 10#
#3x - 2x = 10#
#x = 10#