What is the answer to 3^x+3^{-x}=12 ?

The answer to 3^x+3^{-x}=12 is x=(ln(6+sqrt(35)))/(ln(3)),x=(ln(6-sqrt(35)))/(ln(3))

Solve for x: 3^x+3^{-x}=12

The solution is x=(ln(6+sqrt(35)))/(ln(3)),x=(ln(6-sqrt(35)))/(ln(3))

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3^{x} + 3^{- x} = 12

x = \frac{ln ( 6 + 35 )}{ln ( 3 )}, x = \frac{ln ( 6 - 35 )}{ln ( 3 )}

Decimal Notation

x = 2.25547 \dots, x = - 2.25547 \dots

Solution steps

3^{x} + 3^{- x} = 12

Apply exponent rules

3^{x} + (3^{x})^{- 1} = 12

Rewrite the equation with

3^{x} = u

u + (u)^{- 1} = 12

Solve

u + u^{- 1} = 12 : u = 6 + 35, u = 6 - 35

u = 6 + 35, u = 6 - 35

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 6 + 35 : x = \frac{ln ( 6 + 35 )}{ln ( 3 )}

Solve

3^{x} = 6 - 35 : x = \frac{ln ( 6 - 35 )}{ln ( 3 )}

x = \frac{ln ( 6 + 35 )}{ln ( 3 )}, x = \frac{ln ( 6 - 35 )}{ln ( 3 )}

3^{y} + 3^{1 - y} = 4

7 = 3^{x}

3^{3 n} = 4^{3 n - y}

a, \frac{4 ^{a + 1}}{b} = 125

8^{x} = 32

What is the answer to 3^x+3^{-x}=12 ?
The answer to 3^x+3^{-x}=12 is x=(ln(6+sqrt(35)))/(ln(3)),x=(ln(6-sqrt(35)))/(ln(3))
Solve for x: 3^x+3^{-x}=12
The solution is x=(ln(6+sqrt(35)))/(ln(3)),x=(ln(6-sqrt(35)))/(ln(3))