Solve for x log2 x + log4 x + log8 x =11/6
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We have to solve for x: log2 x + log4 x + log8 x =11/6
log (a) b = 1/ log (b) a
log2 x + log4 x + log8 x =11/6
=> 1 / log(x) 2 + 1 / log(x) 4 + 1/log(x) 2 = 11/6
=> 1 / log(x) 2 + 1 / 2*log(x) 2 + 1/3*log(x) 2 = 11/6
let log(x) 2 = y
=> 1/y + 1/2y + 1/3y = 11/6
=> 6/6y + 3/6y + 2/6y = 11/6
=> 11/6y = 11/6
=> y = 1
So log(x) 2 = 1
=> 2 = x^1
=> x = 2
Therefore x = 2
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First, we'll impose the constraints of existence of logarithms:
x>0
Now, we'll change the bases of logarithms to base 2:
log2 x = log4 x*log2 4
log4 x = log2 x/log2 (2^2)
log4 x = log2 x/2log2 2
But log2 2 = 1
log4 x = log2 x/2
log2 x = log8 x*log2 8
log8 x = log2 x/3log2 2
log8 x = log2 x/3
We'll re-write the equation
log2 x + (log2 x)/2 + (log2 x)/3 = 11/6
We'll multiply by 6 both sides:
6 log2 x + 3 log2 x + 2 log2 x = 11
We'll add like terms:
11 log2 x = 11
We'll divide by 11:
log2 x = 1
x = 2^1
x = 2
Since 2 belongs to the interval of admissible values, we'll validate it as solution of the equation:
x = 2
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