How do you solve #log_2(x+7)+ log_2 (x+8)=1#?

1 Answer
Gió
Nov 30, 2015

I found #x=-6#

Explanation:

You can use the property of the logs that tells us that:
#logx+logy=log(xy)#
and write:
#log_2[(x+7)(x+8)]=1#
now you apply the definition of log to change it into an exponential:
#[(x+7)(x+8)]=2^1#
#[(x+7)(x+8)]=2#
so that:
#x^2+8x+7x+56=2#
#x^2+15x+54=0#
We can use the Quadratic Formula:
#x_(1,2)=(-15+-sqrt(225-216))/2=(-15+-3)/2#
Two solutions:
#x_1=-9# NO it makes both argument negative.
#x_2=-6#

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