What are x values? log_x_2 + log_2x_2= log_4x_2 x,2x,4x are bases of logarithms
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We have to solve the equation: log(x) 2 + log(2x) 2 = log(4x) 2
Now log(a) b = 1/ log (b) a
So we can write
log(x) 2 + log(2x) 2 = log(4x) 2
=> 1/log (2) x + 1/ log 2( 2x) = 1/ log 2 ( 4x)
=> 1/ log (2) x + 1/[ log (2) x + log 2 (2)] = 1/ [ log(2)x + log(2) 2^2]
=> 1/ log(2)x + 1/[log(2)x + 1] = 1/[log(2)x + 2]
Let log(2)x = y
=> 1/y + 1/(y +1) = 1/(y+2)
=> [y+1 +y]/y*(y+1) = 1/(y+2)
=> (2y+1)(y+2) = y(y+1)
=> 2y^2 + 4y +y +2 = y^2 + y
=> y^2 + 4y + 2 = 0
y1 = [-4 + sqrt (16 - 8)]/2
=> [-2 + sqrt 2]
y2 = -2 - sqrt 2
y = log (2) x
log(2)x = -2 + sqrt 2
=> x = 2^(-2 + sqrt 2)
and x = 2^(-2 - sqrt 2)
Therefore x = 2^(-2 + sqrt 2) and 2^(-2 - sqrt 2)
Related Questions
First, we'll impose the constraints of existence of logarithms:
x>0 and x different from 1
2x>0 and x different from 1/2
4x > 0 and x different from 1/4
Now, we'll solve the equation interchanging the bases and arguments at each of the terms of the given equation.
1/log2 x + 1/log2 2x = 1/log2 4x
We'll apply the product property of logarithms:
log2 2x = log2 2 + log2 x
But log2 2 = 1
log2 2x = 1 + log2 x
log2 4x = log2 4 + log2 x
log2 4x = log2 2^2 + log2 x
log2 4x = 2log2 2 + log2 x
log2 4x = 2 + log2 x
We'll re-write the equation:
1/log2 x + 1/(1 + log2 x) = 1/(2 + log2 x)
We notice the presence of log2 x all over the equation, so, we'll substitute it by t.
1/t + 1/(1+t) = 1/(2+t)
We'll multiply each terms by t(1+t)(2+t):
t(1+t)(2+t)/t + t(1+t)(2+t)/(1+t) = t(1+t)(2+t)/(2+t)
We'll simplify and we'll get:
(1+t)(2+t) + t(2+t) = t(1+t)
We'll remove the brackets using FOIL:
2 + 3t + t^2 + 2t + t^2 = t + t^2
We'll subtract t + t^2:
2 + 3t + t^2 + 2t + t^2 - t - t^2 = 0
We'll combine like terms:
t^2 + 4t + 2 = 0
We'll apply quadratic formula:
t1 = [-4+sqrt(16 - 8)]/2
t1 = -2+sqrt2
t2 = -2-sqrt2
But log2 x = t
log2 x = -2+sqrt2
x1 = 2^(-2+sqrt2)
x2 = 1/2^(2+sqrt2)
The solutions of the equation are: x1 = 2^(-2+sqrt2) and x2 = 1/2^(2+sqrt2).
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