What is the answer to (3)^{2x+1}+3^2=28(3^x) ?

The answer to (3)^{2x+1}+3^2=28(3^x) is x=2,x=-1

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(3)^{2x+1}+3^2=28(3^x)

(3)^{2 x + 1} + 3^{2} = 28 (3^{x})

x = 2, x = - 1

Solution steps

(3)^{2 x + 1} + 3^{2} = 28 (3^{x})

Apply exponent rules

(3^{x})^{2} \cdot 3^{1} + 3^{2} = 28 \cdot 3^{x}

Rewrite the equation with

3^{x} = u

(u)^{2} \cdot 3^{1} + 3^{2} = 28 u

Solve

u^{2} \cdot 3^{1} + 3^{2} = 28 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

5000 = 1000 (1.07)^{n}

16 = 2^{t}

3^{a} = 81

2 7^{- 3 x + 2} = 81

e^{(x)} - 5 e^{(- x)} + 4 e^{(- 3 x)} = 0

What is the answer to (3)^{2x+1}+3^2=28(3^x) ?
The answer to (3)^{2x+1}+3^2=28(3^x) is x=2,x=-1
Solve for x: (3)^{2x+1}+3^2=28(3^x)
The solution is x=2,x=-1