Can anybody please solve it?
![enter image source here]()
answers are x=2 or #1-root2 33#
I got x=2 but did not get x=#1-root2 33#
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"Question #99ee1"
#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#.
#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#
