# (log 2 x^2)^2 + log 2 x -18 = 0

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

l(og 2 x^2 )^2 + log 2 x - 18 = 0

==> (2log 2 (x))^2 + log 2 x -18 = 0

==> 4(log 2 (x))^2 + log 2 x  -18 = 0

Let us assume that y= log 2 (x)

==> 4y^2 + y - 18 = 0

==> y1= [-1+sqrt(1+4*4*18)]/8 = -1+17/8 = 16/8 = 2

==> y1= 2  ==> log 2 (x1) = y1= 2

==> x1= 4

==> y2= -1-17/8 = -18/8 = -9/4

==> y1= -9/4 ==> log 2 (x2) = -9/4 ==> x2= 2^(-9/4)

==>  x2= (2)^(-9/4)

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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It is a logarithmic equation and we'll solve it using substitution technique.

First, we'll use the power property for the first term of the equation:

(log 2 x^2)^2 = 2*(log 2 x)^2

Now, we'll substitute log 2 x = t.

The equation will become:

2t^2 + t - 18 = 0

We'll apply the quadratic formula:

t1 = [-1+sqrt(1+144)]/4

t1 = (-1+sqrt145)/4

t2 = (-1- sqrt145)/4

Now, we'll calculate x:

log 2 x = t1

log 2 x = (-1+sqrt145)/4

x = 2^ (-1+sqrt145)/4

log 2 x = (-1-sqrt145)/4

x = 2^(-1-sqrt145)/4

x = 1/2^(1+sqrt145)/4

tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The equation `(log_2 x^2)^2 + log_2 x -18 = 0` has to be solved

`(log_2 x^2)^2 + log_2 x -18 = 0`

Use the property of logarithm, log a^b = b*log a. This gives:

`(2*log_2 x)^2 + log_2 x -18 = 0`

Now take the square of the first term

`2^2*(log _2x)^2 + log_2 x -18 = 0`

Let `y = log_ 2x`

4y^2 + y - 18 = 0

4y^2 + 9y - 8y - 18 = 0

y(4y + 9) - 2(4y + 9) = 0

(y - 2)(4y + 9) = 0

y = 2 and y = -9/4

Now `y = log_2 x`

`log_2 x = 2`

x = 2^2 = 4

`log_2 x = -9/4`

x = 2^(-9/4) which is approximately 0.21022

The solution of the equation is x = 4 and x = 0.21022

jeyaram | Student, Undergraduate | (Level 1) Valedictorian

Posted on

﻿(log 2 x^2)^2 + log 2 x -18 = 0

substitute log 2 x = t.

4t^2+t-18=0

(t-2)(4t+9)=0

t=2    or       t=(-9/4)

log 2 x=2     or     log 2 x=(-9/4)

x=4   or    x=0.210

neela | High School Teacher | (Level 3) Valedictorian

Posted on

(log2x^2)^2 +log2x -18 = 0. To solve for x.

Solution:

We write the equation  under base 2:

(logx^2)^2 +logx -18 = 0, The base is 2.

(2logx)^2 +logx -18 = 0

4(logx)^2 +logx -18.

4y^2 +y -18 = 0, where u = logx base 2

4y^2 +9y -8y-18 = 0

y(4y+9) - 2(4y+9) = 0

(4y+9)(y-2) = 0

4y+9 = 0 or y -2 = 0 .

y-2 = 0 gives: log 2 (x) = 2

x = 2^2 =4.

4y + 9 = 0 gives: y  log2(x)= -9/4 which gives no real solution .

So x = 4 is the real solution.

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