How do you solve for x in #Log x + Log (3x-13)=1#?

Question

5733 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-20T04:23:32

#x=5# only

If the question just says #log#, we can safely assume that it means #log_10#

#log_10x+log_10(3x-13)=1#

#log_10 x(3x-13)=1#
Recall: #log_a b+log_a c=log_a bc#

#x(3x-13)=10^1#
Recall: If #log_a b=c# then #a^c=b#

#x(3x-13)=10#

#3x^2-13x-10=0#

#(3x+2)(x-5)=0#

#x=-2/3# or #x=5#

But #x=-2/3# is not a solution since you cannot log negative numbers so #x=5# is the solution only