Logarithm question

Question

Log_(1/3)(1+x)/8= log_(1/3)3/(x+6)

I need to find the domain and solution

Answer 1

The domain is simple. You cannot take the log of numbers less than or equal to zero. So (1+x)/8 >0 and 3/(x+6)>0

For (1+x)/8 >0 x must be > -1

For 3/(x+6)>0 x must be > -6

The top one is more restrictive than the bottom so the domain is x > -1 or (-1, ∞)

The solution is found by just dropping the logs since both sides are log(1/3)

So (1+x)/8 = 3/(x+6)

Cross multiplying: (x + 1)(x + 6) = (8)(3)

x² + 7x + 6 = 24

x² + 7x - 18 = 0

(x + 9)(x - 2) = 0

x = -9 or x = 2

We can throw out x = -9 since that is not in the domain so the solution is x = 2

Answer 2

The domain of log function is all real number.

for example if you have a log(x+2) , your condition to find the domain is to set (x+2) > 0 and solve for x.

In your case you have two logs , so you need to identify what are the possible values for each log to satisfy the condition.

(1+x)/8 > 0 => , x >-1;

3/(x+6) > 0. => you have all real number except -6 .

Therefore your domain is (-1, infinity)

Solving the equation , you have one to one identity so you set (1+x)/8 = 3/(x+6) => (1+x)(x+6)=24

then solve for x .

x^2+7x+6 =24

x^2 +7x -18 = 0

solve for x and then you check which solution is within your domain.

Hope this helps.