What is the answer to 3^x+3^{2x+1}=4 ?

The answer to 3^x+3^{2x+1}=4 is x=0

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3^x+3^{2x+1}=4

3^{x} + 3^{2 x + 1} = 4

x = 0

Solution steps

3^{x} + 3^{2 x + 1} = 4

Apply exponent rules

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

Rewrite the equation with

3^{x} = u

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

Solve

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

u = 1, u = - \frac{4}{3}

Substitute back

u = 3^{x},

solve for

x

Solve

3^{x} = 1 : x = 0

Solve

3^{x} = - \frac{4}{3} :

No Solution for

x \in R

x = 0

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

5^{x + 1} + 5 + 5^{x - 1} = \frac{31}{5}

\frac{6 ^{x} - 7 ( 6 ^{- x} )}{2} = 3

5^{x} = 2 5^{10 x - 8}

2^{x^{2} + 7 x} = 8^{- 4}