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