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

The answer to 3^{2x}-3(3^x)=4 is x=(2ln(2))/(ln(3))

Solve for x: 3^{2x}-3(3^x)=4

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

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3^{2 x} - 3 (3^{x}) = 4

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

Decimal Notation

x = 1.26185 \dots

Solution steps

3^{2 x} - 3 (3^{x}) = 4

Apply exponent rules

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

Rewrite the equation with

3^{x} = u

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

Solve

u^{2} - u \cdot 3^{1} = 4 : u = 4, u = - 1

u = 4, u = - 1

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 4 : x = \frac{2 ln ( 2 )}{ln ( 3 )}

Solve

3^{x} = - 1 :

No Solution for

x \in R

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

3^{x + 1} = 9^{x - 2}

\frac{A}{1 + B e ^{\frac{t}{2}}} = C

lo g_{2} (2^{x}) = x

2 e^{- x^{2}} = 0

2^{x + 1} + 2^{x} + 2^{x - 2} = 26

What is the answer to 3^{2x}-3(3^x)=4 ?
The answer to 3^{2x}-3(3^x)=4 is x=(2ln(2))/(ln(3))
Solve for x: 3^{2x}-3(3^x)=4
The solution is x=(2ln(2))/(ln(3))