# Solve for x log 3 (x^2+8)-log 3 (4)=3.

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You need to use logarithmic identities, such that:

`log_3 (x^2 + 8) - log_3 4 = log_3 (x^2+8)/4`

`log_3 (x^2 + 8) - log_3 4 = 3 => log_3 (x^2+8)/4 = 3=> (x^2+8)/4 = 3^3 => x^2+8 = 4*27 => x^2 + 8 = 108`

`x^2 = 108 - 8 => x^2 = 100 => x^2 - 100 = 0`

Converting the difference of squares into a product, yields:

`(x - 10)(x + 10) = 0 => {(x - 10 = 0),(x + 10 = 0):}`

`{(x = 10),(x = -10):}`

You need to test the values x = 10 and x = -10 in equation, such that:

`log_3 ((10)^2+8)/4 = 3 => log_3 108/4 = 3 => log_3 27 = 3 => 27 = 3^3 => 27 = 27`

`log_3 ((-10)^2+8)/4 = 3 => log_3 108/4 = 3 => log_3 27 = 3 => 27 = 3^3 => 27 = 27`

**Hence, evaluating the solutions to the given equation yields**` x = -10, x = 10.`

The equation `log_3 (x^2+8)-log_3 (4)=3` has to be solved for x.

Use the property of logarithm `log a - log b = log(a/b)` and `log_a b = c` implies `b = a^c` .

`log_3 (x^2+8)-log_3 (4)=3`

`log_3 ((x^2+8)/4) = 3`

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

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

`x^2 + 8 = 108`

`x^2 = 100`

`x = +- 10`

The solution of the equation `log_3 (x^2+8)-log_3 (4)=3` is `+-10`

before solving the problem, we'll impose the conditions 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

Therefore, the positive solution of the equation is x = 10.