Solve for x :- 1) x^(-3) + x^(-3/2) = 2 2) 6(4^x + 9^x) = 6^x(4+9)
- print Print
- list Cite
Expert Answers
calendarEducator since 2008
write3,662 answers
starTop subjects are Math, Science, and Social Sciences
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
Related Questions
- Solve for x (x^2-9)^1/2=4
- 1 Educator Answer
- solve for x if x^3 +x^2 +x +1 = 0
- 2 Educator Answers
- Solve algebraically for x: (x + 2)/6 = 3/(x - 1)
- 2 Educator Answers
- Solve for x: √3 sin x + cosx = 1
- 1 Educator Answer
- 5/6·1/2+2/3÷4/3 How do I solve this problem?
- 1 Educator Answer
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.
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.
Student Answers