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Solve for x :- 1) x^(-3) + x^(-3/2) = 2 2) 6(4^x + 9^x) = 6^x(4+9)

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rnk | Student, Grade 10 | (Level 1) Honors

Posted June 7, 2010 at 3:23 PM via web

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Solve for x :-

1) x^(-3) + x^(-3/2) = 2

2) 6(4^x + 9^x) = 6^x(4+9)

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neela | High School Teacher | (Level 3) Valedictorian

Posted June 7, 2010 at 4:25 PM (Answer #1)

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1) to solve x^(-3)+x^(-3/2) = 2.

Put x^(-3/2) =t.Then x^(-3) = t^2.So, the given equation transforms:

t^2+t=2

t^2+t-2 = 0.

(t+2)(t-1) = 0. So , t+2 = 0 or t-1 = 0. Or t =-2. Or t =1.

t=-2 implies x^(-3/2) = -2 Or x^(-3) = (-2)^2 =4 . So x^3 = 1/4 Or x = (1/4)^(1/3) = 2^(-2/3)

t=1 gives x^(-3/2) = 1. Or x = 1^(-2/3)= 1.

2)

6(4^x+9^x) = 6^x(4+9).

(4^x+6^x)/(6^x) = 13/6.

(2/3)^x+ (3/2)^x = 4/6+9/6 = (2/3)^1 +(3/2)^1.

So x= 1. Or x = -1

Aliter:

(2/3)^x+(3/2)^x = 13/6. Put (2/3)^x = a, Then (3/2 )^x = 1/a. So the equation transforms as below:

a+1/a = 13/6 . Multiplying by 6a, both sides, we get:

6a^2+6=13a

6a^2-13a+6=0

6a^2-9a-4a+6 = 0.

3a(2a-3)-2(2a-3) = 0.

(2a-3)(3a-2) = 0. Or

2a=3 or 3a=2. Or

a=3/2 Or a = 2/3. but a = (2/3)^x.

So (2/3)^x = 3/2 gives x = -1

(2/3)^x = (2/3) gives x = 1.

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

Posted June 7, 2010 at 10:21 PM (Answer #2)

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1. x^-3 +x^-3/2 = 2

assume x^-3=y

\==> y+y^1/2 =2

==> y + y^1/2 -2=0

==> (y^1/2 +2)(y^1/2 -1)=0

==> y^1/2= -2 which is impossible

==> OR  y^1/2= 1 ==> y=1 ==> x= 1

2.   6(4^x + 9^x)= 6^x(9+6)

==> 6(4^x +9^x) = 6^x (13)

Divide by 6(6^x)

==> (4^x + 9^x)/6^x= 13/6

==> 4^x/6^x  + 9^x/6^x = 13/6

==> (2/3)^x + (3/2)^x = 13/6

Now assume that (2/3)^x =y

==> y+ 1/y = 13/9

==> y^2 + 1 = (13/9)y

==> y^2-(13/9)y +1 =0

==> [y-(3/2)][y-(2/3)]=0

==> y = 3/2 ==> x= -1

 OR y= 2/3 ==> x=1

 

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted June 7, 2010 at 10:26 PM (Answer #3)

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For solving the problem no. 2, I'll suggest the way of solving:

First, we'll open the left bracket and we'll get:

6*4^x + 6*9^x = 13*6^x

We'll write: 4^x=2^2x, 9^x=3^2x and 6^x=(2*3)^x

6*2^2x + 6*3^2x - 13*(2*3)^x = 0

Because 3^2x is never cancelling, we'll divide entire equation by 3^2x:

6*(2/3)^2x - 13*(2/3)^x + 6 = 0

We'll note (2/3)^x=t

6*t^2 - 13*t + 6 = 0

We'll use the quadratic formula:

t1 = {13+sqrt[(13-12)(13+12)]}/12

t1 = (13+5)/12

t1 = 18/12

t1 = 3/2

t2 = (13-5)/12

t2 = 8/12

t2 = 2/3

Let's recall that we didn't find out x yet.

So (2/3)^x = t1

(2/3)^x = 3/2

(2/3)^x = 1/(3/2)

(2/3)^x = (2/3)^-1

We'll use one to one law and we'll get:

x=-1

We'll do the same for (2/3)^x = t2

(2/3)^x = 2/3

Again we'll use the one to one law:

x=1

So, the solutions for the equation are:

x1=1 and x2=-1.

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