Question

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

Answer 1

Split the addition of exponents into multipliciation:

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

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

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

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

Let `u=3^x`:

`3u^2-12u+9=0`

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

`(u-1)(u-3)=0`

`u=1,3`

Put `3^x` back in for `u`:

#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!

Answer 2

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`

Answer 3

`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`