What is the answer to 3^{2x-2}+3^{x+1}=12 ?

The answer to 3^{2x-2}+3^{x+1}=12 is x=(ln((9(-3+sqrt(43/3)))/2))/(ln(3))

Solve for x: 3^{2x-2}+3^{x+1}=12

The solution is x=(ln((9(-3+sqrt(43/3)))/2))/(ln(3))

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3^{2 x - 2} + 3^{x + 1} = 12

x = \frac{ln ( \frac{9 ( - 3 + \frac{43}{3} )}{2} )}{ln ( 3 )}

Solution steps

3^{2 x - 2} + 3^{x + 1} = 12

Apply exponent rules

(3^{x})^{2} \cdot 3^{- 2} + 3^{x} \cdot 3^{1} = 12

Rewrite the equation with

3^{x} = u

(u)^{2} \cdot 3^{- 2} + u \cdot 3^{1} = 12

Solve

u^{2} \cdot 3^{- 2} + u \cdot 3^{1} = 12 : u = \frac{9 ( - 3 + \frac{43}{3} )}{2}, u = \frac{9 ( - 3 - \frac{43}{3} )}{2}

u = \frac{9 ( - 3 + \frac{43}{3} )}{2}, u = \frac{9 ( - 3 - \frac{43}{3} )}{2}

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = \frac{9 ( - 3 + \frac{43}{3} )}{2} : x = \frac{ln ( \frac{9 ( - 3 + \frac{43}{3} )}{2} )}{ln ( 3 )}

Solve

3^{x} = \frac{9 ( - 3 - \frac{43}{3} )}{2} :

No Solution for

x \in R

x = \frac{ln ( \frac{9 ( - 3 + \frac{43}{3} )}{2} )}{ln ( 3 )}

2^{2 x - 1} - 5 \cdot 2^{x + 1} + 32 = 0

1^{x} = 0

1^{x} = 1

4^{z} = 8

2^{x} = 4096

What is the answer to 3^{2x-2}+3^{x+1}=12 ?
The answer to 3^{2x-2}+3^{x+1}=12 is x=(ln((9(-3+sqrt(43/3)))/2))/(ln(3))
Solve for x: 3^{2x-2}+3^{x+1}=12
The solution is x=(ln((9(-3+sqrt(43/3)))/2))/(ln(3))