What is the answer to 1+9^x=3^{x+1}+3^{x-1} ?

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

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1+9^x=3^{x+1}+3^{x-1}

1 + 9^{x} = 3^{x + 1} + 3^{x - 1}

x = 1, x = - 1

Solution steps

1 + 9^{x} = 3^{x + 1} + 3^{x - 1}

Simplify

3^{x + 1} + 3^{x - 1} : 10 \cdot 3^{x - 1}

1 + 9^{x} = 10 \cdot 3^{x - 1}

Apply exponent rules

1 + (3^{x})^{2} = 10 \cdot 3^{x} \cdot 3^{- 1}

Rewrite the equation with

3^{x} = u

1 + (u)^{2} = 10 u \cdot 3^{- 1}

Solve

1 + u^{2} = 10 u \cdot 3^{- 1} : u = 3, u = \frac{1}{3}

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

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 3 : x = 1

Solve

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

x = 1, x = - 1

\frac{9 ^{x}}{9 ^{3}} = 9^{5}

6^{x} = 12

3^{x + 1} = 8

1 9^{x} = 0

2^{2 x} - 5 \cdot 2^{x + 2} + 64 = 0

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