How do I prove that x=1x=1 or 00 from solving 3^(2x+1) - 12(3^x) + 9 = 032x+112(3x)+9=0?

3 Answers
Apr 4, 2018

Split the addition of exponents into multipliciation:

3^(color(red)(2x)+color(blue)1)-12(3^x)+9=032x+112(3x)+9=0

3^color(red)(2x)*3^color(blue)1-12(3^x)+9=032x3112(3x)+9=0

3(3^color(red)(2x))-12(3^x)+9=03(32x)12(3x)+9=0

3(3^color(red)x)^2-12(3^x)+9=03(3x)212(3x)+9=0

Let u=3^xu=3x:

3u^2-12u+9=03u212u+9=0

u^2-4u+3=0u24u+3=0

(u-1)(u-3)=0(u1)(u3)=0

u=1,3u=1,3

Put 3^x3x back in for uu:

color(white){color(black)( (3^x=color(red)1,qquad3^x=color(red)3), (3^x=color(red)(3^0),qquad3^x=color(red)(3^1)), (x=color(red)0,qquadx=color(red)1):}

That's it. Hope this helped!

Apr 4, 2018

See below

Explanation:

3^(2x+1)-12·3^x+9=0 Reorganize

3·3^(2x)-12·3^x+9=0

lets make a change z=3^x. Then we have

3z^2-12z+9=0 use the cuadratic formula

z=(12+-sqrt(12^2-4·9·3))/6=(12+-sqrt(144-108))/6=(12+-6)/6

The we have z_1=3 and z_2=1

Undo the change 3^x=z_1=3, then x=1

3^x=z_2=1, then x=0

Apr 4, 2018

x=1 or x=0

Explanation:

Here,

3^(2x+1) - 12(3^x) + 9 = 0.....to[ as ,color(blue)(a^(m+n)=a^m*a^n]

3^(2x)3^1 - 12(3^x) + 9 = 0

3(3^x)^2-12(3^x)+9=0...to[as,color(blue)((a^m)^n=a^(mn)]

Dividing both sides by 3,we get

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

Let , 3^x=a

a^2-4a+3=0

We have,

(-3)+(-1)=-4, and

(-3)xx(-1)=3

So,

a^2color(red)(-3a-1a)+3=0

a(a-3)-1(a-3)=0

(a-3)(a-1)=0

=>a-3=0 or a-1=0

=>a=3 or a=1.....towhere, color(blue)(a=3^x

3^x=3^1 or 3^x=3^0

=>x=1 or x=0