What is the answer to 3^{2x-2}+7=4(3^{x-1}+1) ?

The answer to 3^{2x-2}+7=4(3^{x-1}+1) is x=2,x=1

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3^{2x-2}+7=4(3^{x-1}+1)

3^{2 x - 2} + 7 = 4 (3^{x - 1} + 1)

x = 2, x = 1

Solution steps

3^{2 x - 2} + 7 = 4 (3^{x - 1} + 1)

Apply exponent rules

(3^{x})^{2} \cdot 3^{- 2} + 7 = 4 (3^{x} \cdot 3^{- 1} + 1)

Rewrite the equation with

3^{x} = u

(u)^{2} \cdot 3^{- 2} + 7 = 4 (u \cdot 3^{- 1} + 1)

Solve

u^{2} \cdot 3^{- 2} + 7 = 4 (u \cdot 3^{- 1} + 1) : u = 9, u = 3

u = 9, u = 3

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 9 : x = 2

Solve

3^{x} = 3 : x = 1

x = 2, x = 1

y, 17 y e^{- x^{2} - y^{2}} - 34 x^{2} y e^{- x^{2} - y^{2}} = 0

2^{x + 6} = 2^{5}

9^{x} - 8 \times 3^{x} - 9 = 0

(5^{x - 2})^{x - 3} = 1

e^{2 x} x = 2

What is the answer to 3^{2x-2}+7=4(3^{x-1}+1) ?
The answer to 3^{2x-2}+7=4(3^{x-1}+1) is x=2,x=1
Solve for x: 3^{2x-2}+7=4(3^{x-1}+1)
The solution is x=2,x=1