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

The answer to 3^{2x+1}+3^2=3^{x+3}+3^x is x=2,x=-1

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

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

x = 2, x = - 1

Solution steps

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

Apply exponent rules

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

Rewrite the equation with

3^{x} = u

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

Solve

u^{2} \cdot 3^{1} + 3^{2} = u \cdot 3^{3} + u : u = 9, u = \frac{1}{3}

u = 9, u = \frac{1}{3}

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 9 : x = 2

Solve

3^{x} = \frac{1}{3} : x = - 1

x = 2, x = - 1

2^{x} = - 1

3.6 = (1.025)^{n}

6 e^{3 x} - 18 e^{- 2 x} = 0

2^{x} = 10

2^{x} = 12

What is the answer to 3^{2x+1}+3^2=3^{x+3}+3^x ?
The answer to 3^{2x+1}+3^2=3^{x+3}+3^x is x=2,x=-1
Solve for x: 3^{2x+1}+3^2=3^{x+3}+3^x
The solution is x=2,x=-1