Log 3 ( x^2 + 8) - log 3 (4) = 3

We will use algorithm properties to solve:

We know that:

log a (b) - log a (c) = log a (b/c)

==> log 3 (x^2+8)/(4) = 3

Now we will rewrite using exponent:

==> (x^2 + 8)/4 = 3^3

==> (x^2+8)/4 = 27

Now multiply by 4:

==> (x^2+ 8) = 27*4

==> x^2 + 8 = 108

Now we will subtract 8 from both sides:

==> x^2 = 100

**==> x= +-10**

First, we'll impose the constraint of existence of logarithms:

x^2 + 8 > 0 (is positive for any value of x)

x>0 => x belongs to the interval (0, +infinite)

Now, we'll solve the equation:

Log 3 ( x^2 + 8) - log 3 (4) = 3

We'll add log 3 (4) both sides:

Log 3 ( x^2 + 8) = log 3 (4) + 3

We'll write 3 = 3* log 3 (3)

We'll use the power rule of logarithms:

3* log 3 (3) = log 3 (3^3)

3* log 3 (3) = log 3 (27)

We'll re-write the equation:

Log 3 ( x^2 + 8) = log 3 (4) + log 3 (27)

We'll use the product rule of logarithms:

log a + log b = log a*b

log 3 (4) + log 3 (27) =log 3(4*27)

log 3 (4) + log 3 (27) = log 3 (108)

The equation will become:

Log 3 ( x^2 + 8) = log 3 (108)

Since the bases are matching, we'll get:

x^2 + 8 = 108

x^2 = 108 - 8

x^2 = 100

x = +/- sqrt 100

**x1 = 10**

x2 = -10

Since the second solution has a negative value, we'll reject it and we'll keep just the positive value of the solution.

**So, the solution of the equation is x = 10.**