Question

Can anybody please solve it?

answers are x=2 or `1-root2 33`
I got x=2 but did not get x=`1-root2 33`

Answer 1

`x = 2`

Explanation:

`3/2 log_4(x+2)^2 + 3 = log_4(4-x)^3 + log_4(6+x)^3`

Use `log(a)^x = x log(a)`

`3/cancel{2} cdot cancel{2} log_4(x+2) + 3 = 3 log_4(4-x) + 3log_4(6+x)`

Dividing both sides by 3,

`log_4(x+2) + 1 = log_4(4-x) + log_4(6+x)`

Using `log(x) + log(y) = log(xy)`

`log_4(x+2) + 1 = log_4[(4-x)(6+x)]`
`log_4[24+4x-6x-x^2] - log_4(x+2) = 1`

Use `logx - logy = log(x/y)`

`log_4[ (24 - 2x - x^2)/ (x+2)] = 1 rightarrow text{Definition of log}`

`(24 - 2x - x^2)/(x+2) = 4^1 rightarrow 24 - 2x - x^2 = 4x + 8`

`x^2 + 6x - 16 = 0`

`x_{1,2} = (-6 pm sqrt(36+64))/2 = (-6 pm 10)/2 = {2, -8}`

If we look at the original equation, there's no problem plugging in 2, but if we plug in `x=-8`, we will get the logarithm of a negative number, which is not really allowed (if you use complex numbers it is, but generally we assume otherwise, for simplicity).

Therefore, the only acceptable solution is `x=2`.

Answer 2

`x = {2, 1 - sqrt(33)}`

Explanation:

`3/2 log_4(x+2)^2 + 3 = log_4(4-x)^3 + log_4(6+x)^3`

Since we are taking the power down, we have to consider absolute values. For even exponents, we need them, hence
`3 log_4|x+2| + 3 = 3log_4(4-x) + 3log_4(6+x)`
`log_4 4|x+2| = log_4((4-x)(6+x))`

`4|x+2| = (4-x)(6+x)`

Assuming `x+2 > 0`,
`4(x+2) = (4-x)(6+x) rightarrow x^2 + 6x -16 = 0`
`x = {2, -8}`
As stated above, the `x=-8` solution is invalid because we have to take log of `(6+x)` and that cannot be negative.

Assuming `x+2 < 0`,
`-4(x+2) = (4-x)(6+x) rightarrow x^2 -2x - 32 = 0`
`x = 1 pm sqrt(33)`
Since `x+2 < 0`, the positive solution does not apply.

The limiting condition is that we need to take the logarithm of
`6+x` and since `6 + (1 - sqrt33) > 0`, we are good.

So the other solution is `x = 1 - sqrt33`