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

Apr 4, 2018

Split the addition of exponents into multipliciation:

${3}^{\textcolor{red}{2 x} + \textcolor{b l u e}{1}} - 12 \left({3}^{x}\right) + 9 = 0$

${3}^{\textcolor{red}{2 x}} \cdot {3}^{\textcolor{b l u e}{1}} - 12 \left({3}^{x}\right) + 9 = 0$

$3 \left({3}^{\textcolor{red}{2 x}}\right) - 12 \left({3}^{x}\right) + 9 = 0$

$3 {\left({3}^{\textcolor{red}{x}}\right)}^{2} - 12 \left({3}^{x}\right) + 9 = 0$

Let $u = {3}^{x}$:

$3 {u}^{2} - 12 u + 9 = 0$

${u}^{2} - 4 u + 3 = 0$

$\left(u - 1\right) \left(u - 3\right) = 0$

$u = 1 , 3$

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!

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

$3 {z}^{2} - 12 z + 9 = 0$ 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

$x = 1 \mathmr{and} x = 0$

#### Explanation:

Here,

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

${3}^{2 x} {3}^{1} - 12 \left({3}^{x}\right) + 9 = 0$

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

${\left({3}^{x}\right)}^{2} - 4 \left({3}^{x}\right) + 3 = 0$

Let , ${3}^{x} = a$

${a}^{2} - 4 a + 3 = 0$

We have,

$\left(- 3\right) + \left(- 1\right) = - 4$, and

$\left(- 3\right) \times \left(- 1\right) = 3$

So,

${a}^{2} \textcolor{red}{- 3 a - 1 a} + 3 = 0$

$a \left(a - 3\right) - 1 \left(a - 3\right) = 0$

$\left(a - 3\right) \left(a - 1\right) = 0$

$\implies a - 3 = 0 \mathmr{and} a - 1 = 0$

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

${3}^{x} = {3}^{1} \mathmr{and} {3}^{x} = {3}^{0}$

$\implies x = 1 \mathmr{and} x = 0$