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# 32x+3x+1=5-3x

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32x+3x+1=5-3x

Posted by banarsi on July 1, 2010 at 5:41 PM via web and tagged with 111, math

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To solve 3^(2x)+3^(x+1) = 5-3^x.

Solution:

3^2x = (3^x)^2

3^(x+1) = 3*3^x

Putting 3^x = t the given equation becomes:

t^2+3t =5-t

t^2+3t+t -5 = 0

t^2+4t-5 = 0

t^2+5t - t - 5 = 0.

t(t+5) -1(t+5) = 0

(t+5)(t-1) = 0.

t+5 = 0 or t-1 = 0.

t=-5, t =1

t= -5  gives : 3^x = -5 , has no real solution for x.

t = 1 gives: 3^x =1 = 3^0.  x = 0.

Posted by neela on July 1, 2010 at 6:14 PM (Answer #1)

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3^2x + 3^(x+1) = 5-3^x

Let us simplify :

(3^x)^2 + (3)^x (3)^1 = 5- 3^x

(3^x)^2 + 3(3)^x + 3^x -5 =0

(3^x)^2 + 4(3)^x -5 =0

Now assume that 3^x =y

==> y^2 + 4y -5=0

Factorize:

==> (y+5)(y-1)

==> y1= -5 ==> 3^x1=-5  (impossible)

==> y2= 1 ==> 3^x2= 1 ==> x2= 0

Then the only solution is x= 0

Posted by hala718 on July 1, 2010 at 11:37 PM (Answer #2)

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3^(2x) + 3^(x+1) = 5 - 3^x

(3^x)^2 + 3^x * 3^1 = 5 - 3^x

let 3^x = a

a^2 + 3a = 5 - a

a^2 + 4a - 5 = 0

a^2 + 5a - a - 5 = 0

a(a + 5) - (a + 5) = 0

(a - 1)(a + 5) = 0

a = 1 or a = -5

3^x = 1 or 3^x = -5

since (3^x = -5) is impossible

we will take only

3 ^x = 1

We see that this equation satisfies only when x is 0

therefore x = 0

Posted by rnk on July 1, 2010 at 7:35 PM (Answer #3)