# Solve log3 (x+3)= 3-log3 (3x+1)?could you please show me the steps of answering the question thanks

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### 2 Answers

`log_3 (x+3) = 3 - log_3 (3x+1)`

Before we find the root, we will find the domain .

==> x+ 3 > 0 and (3x+1)>0

`==gt x gt -3 and x gt -1/3 `

Then, the domain is `x gt -1/3`

`==gt x in (-1/3, oo).`

Now, we will solve.

First we will move `log_3 (3x+1)` to the left side.

`==gt log_3 (x+3) + log_3 (3x+1) = 3`

Now we will apply logarithm property.

We know that log a + log b = log (ab)

`==gt log_3 (x+3) + log_3 (3x+1) = log_3 (x+3)(3x+1) = 3 `

`==gt log_3 (3x^2 + 10x + 3) = 3`

Now we will rewrite in exponent form.

`==gt 3x^2 + 10x + 3 = 3^3 `

`==gt 3x^2 + 10x + 3 = 27 `

`==gt 3x^2 + 10 x - 24 = 0`

Now we will find the roots.

`==gt x_1= (-10 + sqrt(388))/6 = (-5+sqrt97)/3 ~~ 1.62 `

`==gt x_2= (-5-sqrt97)/3 ~~ -4.95`

But `x_2` in not in the domain.

**Then, the only solution is x = `(-5+sqrt97)/3` **

You need to move all logarithms to the left side, keeping the constant term to the right side such that:

`log_3 (x+3) + log_3 (3x+1) = 3`

You should use the logarithmic property that converts the sum of two logarithms that have like bases in logarithm of product of numbers such that:

`log_3 ((x+3)* (3x+1)) = 3`

`((x+3)* (3x+1)) = 3^3`

You need to open brackets such that:

`3x^2 + x + 9x + 3 - 27 = 0`

`3x^2 + 10x - 24 = 0`

You need to use quadratic formula such that:

`x_(1,2) = (-10+-sqrt(100 + 288))/6`

`x_(1,2) = (-10+-2sqrt97)/6 =gt x_(1,2) = (-5+-sqrt97)/3`

You only need to keep the solution `x_1 = (-5+sqrt97)/3` since if you substitute `(-5+sqrt97)/3` for x in equation, this value checks both logarithms.

**Hence, the solution to the equation is `x = (-5+sqrt97)/3` .**