Solve for x log2 x + log4 x + log8 x =11/6

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justaguide | College Teacher | (Level 2) Distinguished Educator

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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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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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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