# logx 2*log2x 2 - log4x 2*log8x 2=0 x,2x,4x and 8x are in the base

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### 1 Answer

You need to use the following logarithmic identity such that:

`log_a b = 1/(log_b a)`

Reasoning by analogy yields:

`log_x2 = 1/(log_2 x)`

`log_(2x) 2 = 1/(log_2 (2x))`

`log_(4x) 2 = 1/(log_2 (4x))`

`log_(8x) 2 = 1/(log_2 (8x))`

You need to rewrite the equation such that:

`1/((log_2 x)(log_2 (2x))) - 1/((log_2 (4x))*(log_2 (8x)))`

You should use the following logarithmic identity such that:

`log(a*b) = log a + log b`

`(log_2 (2x)) = log_2 2 + log_2 x => (log_2 (2x)) =1 + log_2 x `

`(log_2 (4x)) = log_24 + log_2 x => (log_2 (2x)) =2 + log_2 x `

`(log_2 (8x)) = log_28 + log_2 x => (log_2 (2x)) =3 + log_2 x`

You should come up with the next substitution such that:

`log_2 x = t`

`1/(t(1 + t)) - 1/((2+t)(3+t)) = 0`

You should bring the fraction to a common denominator such that:

`((2+t)(3+t) - t(1+t))/(t(1+t)(2+t)(3+t)) = 0`

Since the denominator needs to be different from zero, hence, you need to solve for t the following equation such that:

`((2+t)(3+t) - t(1+t)) = 0`

`6 + 5t + t^2 - t - t^2 = 0`

Reducing like terms yields:

`4t + 6 = 0 => 4t = -6 => t = -6/4 => t = -3/2`

You need to solve for x the equation `log_2 x = t` such that:

`log_2 x = -3/2 => x = 2^(-3/2) => x = 1/(2^(3/2))`

`x = 1/(2sqrt2) => x = sqrt2/4`

**Hence, evaluating the solution to the given equation yields `x = sqrt2/4.` **

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