How do you solve #log (x + 8) = 1 + log (x - 10)#?

1 Answer
Dec 14, 2015

#x = 12#

Explanation:

Begin by moving both of the #log# terms to the left hand side.

#log(x+8) - log(x-10) = 1#

Now we can use the division rule for logarithms to combine both terms into one. The division rule states that;

#log(m/n) = log(m) - log(n)#

Letting #m=x+8# and #n=x-10#, we get;

#log((x+8)/(x-10)) = 1#

Since we are working with a common #log# it is base ten. That means that the part inside of the parenthesis is equal to #10# raised to the power of the right hand side, or;

#10^1 = (x+8)/(x-10)#

Now we just need to do some algebra to solve for #x#. First, multiply both sides by #(x-10)#.

#10(x-10) = x+8#

Now multiply the #10# through the parenthesis.

#10x - 100 = x + 8#

Subtract #x# and add #100# to both sides.

#9x = 108#

Finally, divide both sides by #9# to find #x#.

#x = 12#