SOLUTION: Can anyone help me solve this problem? log (x^3 + 1) to the base of 3 - log (x+1) to the base of 3 = 1 I tried log to the base of 3 times (x^3 + 1)/(x+1) = 1 (x^3 + 1)/(x+1)

You can put this solution on YOUR website!
Apply the Quotient law to the left side to get the following:
log [[x^3+1]/[x+1]] =1
Factor the sum of cubes and reduce the fraction to get:
log[x^2-x+1]=1
Since the base is "3" you get the following exponential statement:
x^2-x+1=3^1
Simplify and then factor to get:
x^2-x-2=0
(x-2)(x+1)=0
x=2 or x=-1
The solution x=-1 is extraneous. The valid solution is x=2
Cheers,
Stan H.