What is the answer to 3^{2x}-5*3^x+6=0 ?

The answer to 3^{2x}-5*3^x+6=0 is x=1,x=(ln(2))/(ln(3))

Solve for x: 3^{2x}-5*3^x+6=0

The solution is x=1,x=(ln(2))/(ln(3))

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3^{2 x} - 5 \cdot 3^{x} + 6 = 0

x = 1, x = \frac{ln ( 2 )}{ln ( 3 )}

Decimal Notation

x = 1, x = 0.63092 \dots

Solution steps

3^{2 x} - 5 \cdot 3^{x} + 6 = 0

Apply exponent rules

(3^{x})^{2} - 5 \cdot 3^{x} + 6 = 0

Rewrite the equation with

3^{x} = u

(u)^{2} - 5 u + 6 = 0

Solve

u^{2} - 5 u + 6 = 0 : u = 3, u = 2

u = 3, u = 2

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 3 : x = 1

Solve

3^{x} = 2 : x = \frac{ln ( 2 )}{ln ( 3 )}

x = 1, x = \frac{ln ( 2 )}{ln ( 3 )}

3^{x + 1} - 8 (3^{x - 1}) - 5 = 0

\frac{1}{100} = e^{- 0.15 t}

8 (2^{x - 2}) = 1 6^{x}

6^{5 x - 1} - 2 = 1

1 6^{x + 4} = 6 4^{4 x - 5}

What is the answer to 3^{2x}-5*3^x+6=0 ?
The answer to 3^{2x}-5*3^x+6=0 is x=1,x=(ln(2))/(ln(3))
Solve for x: 3^{2x}-5*3^x+6=0
The solution is x=1,x=(ln(2))/(ln(3))