What is the answer to (3^{2x+1}+9)=(3^{x+3}+3^x) ?

The answer to (3^{2x+1}+9)=(3^{x+3}+3^x) is x=2,x=-1

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(3^{2x+1}+9)=(3^{x+3}+3^x)

(3^{2 x + 1} + 9) = (3^{x + 3} + 3^{x})

x = 2, x = - 1

Solution steps

(3^{2 x + 1} + 9) = (3^{x + 3} + 3^{x})

Apply exponent rules

(3^{x})^{2} \cdot 3^{1} + 9 = 3^{x} \cdot 3^{3} + 3^{x}

Rewrite the equation with

3^{x} = u

(u)^{2} \cdot 3^{1} + 9 = u \cdot 3^{3} + u

Solve

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

3^{x} + 3^{x - 1} + 3^{x - 2} = 13

e^{2 x} - 2 e^{x} = 1

1 6^{x} = 8^{2 x + 3}

lo g_{10} (43) = 4. 2^{x}

2^{x} - 2^{2 - x} = 0

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