Can anybody please solve it?

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

2 Answers
Apr 29, 2018

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.

Apr 29, 2018

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