How do you solve #log_10(x+4)-log_10x=log_10(x+2)#?

1 Answer
Jan 30, 2017

I got: #x=(-1+sqrt(17))/2#

Explanation:

We can use a property of logs to write a difference of logs into a log of a fraction:
#log_(10)((x+4)/(x))=log_(10)(x+2)#
If the logs are equal then also the arguments have to be. So:
#(x+4)/x=x+2#
Rearrange and solve for #x#:
#x+4=x^2+2x#
#x^2+x-4=0#
Use the Quadratic Formula:
#x_(1,2)=(-1+-sqrt(1+16))/2#
I can only accept the positive solution to avoid a negative argument in #log_(10)(x)#.