What is the answer to 3^{2x}+3^{x+1}=10 ?

The answer to 3^{2x}+3^{x+1}=10 is x=(ln(2))/(ln(3))

Solve for x: 3^{2x}+3^{x+1}=10

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

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3^{2 x} + 3^{x + 1} = 10

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

Decimal Notation

x = 0.63092 \dots

Solution steps

3^{2 x} + 3^{x + 1} = 10

Apply exponent rules

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

Rewrite the equation with

3^{x} = u

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

Solve

u^{2} + u \cdot 3^{1} = 10 : u = 2, u = - 5

u = 2, u = - 5

Substitute back

u = 3^{x},

solve for

x

Solve

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

Solve

3^{x} = - 5 :

No Solution for

x \in R

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

2^{x} - 4 = - 4^{x} + 4

x, 3^{2 x + 1} + 1 = 3^{x}

43 = 3 (7)^{x}

4^{p} - 4^{p - 1} = 6

x, x e^{y} + y e^{x} = 1

What is the answer to 3^{2x}+3^{x+1}=10 ?
The answer to 3^{2x}+3^{x+1}=10 is x=(ln(2))/(ln(3))
Solve for x: 3^{2x}+3^{x+1}=10
The solution is x=(ln(2))/(ln(3))