SOLUTION: 8. Solve and Check the Following Equations: C. log6 x + log6 (x – 2) = log6 15

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: 8. Solve and Check the Following Equations: C. log6 x + log6 (x – 2) = log6 15      Log On


   

Question 113826: 8. Solve and Check the Following Equations:
C. log6 x + log6 (x – 2) = log6 15
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
log%286%2C%28x%29%29%2Blog%286%2C%28x-2%29%29=log%286%2C%2815%29%29 Start with the given equation


log%286%2C%28x%28x-2%29%29%29=log%286%2C%2815%29%29 Combine the logs using the identity log%28b%2C%28x%29%29%2Blog%28b%2C%28y%29%29=log%28b%2C%28x%2Ay%29%29



log%286%2C%28x%5E2-2x%29%29=log%286%2C%2815%29%29 Distribute


Since the base of the logs are equal, the arguments (the stuff inside the logs are equal). So x%5E2-2x=15


So let's solve x%5E2-2x=15

x%5E2-2x=15 Start with the given equation

x%5E2-2x-15=0 Subtract 15 from both sides


%28x-5%29%28x%2B3%29=0 Factor the left side (note: if you need help with factoring, check out this
solver)



Now set each factor equal to zero:

x-5=0 or x%2B3=0

x=5 or x=-3 Now solve for x in each case


So our possible solutions are x=5 or x=-3


However, since you cannot take the log of a negative number, the only solution is x=5

-------------------------------------
Check:

Let's check the solution x=5

log%286%2C%28x%5E2-2x%29%29=log%286%2C%2815%29%29 Start with the equation found on the third step.


log%286%2C%285%5E2-2%285%29%29%29=log%286%2C%2815%29%29 Plug in x=5



log%286%2C%2825-10%29%29=log%286%2C%2815%29%29 Square and multiply


log%286%2C%2815%29%29=log%286%2C%2815%29%29 Subtract. Since both sides of the equation are equal, this solution is verified.



So the solution x=5 is verified