What is the answer to 3^{x+1}=3^{2x-1}+6 ?

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

Solve for x: 3^{x+1}=3^{2x-1}+6

The solution is x=(ln(6))/(ln(3)),x=1

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3^{x + 1} = 3^{2 x - 1} + 6

x = \frac{ln ( 6 )}{ln ( 3 )}, x = 1

Solution steps

3^{x + 1} = 3^{2 x - 1} + 6

Apply exponent rules

3^{x} \cdot 3^{1} = (3^{x})^{2} \cdot 3^{- 1} + 6

Rewrite the equation with

3^{x} = u

u \cdot 3^{1} = (u)^{2} \cdot 3^{- 1} + 6

Solve

u \cdot 3^{1} = u^{2} \cdot 3^{- 1} + 6 : u = 6, u = 3

u = 6, u = 3

Substitute back

u = 3^{x},

solve for

x

Solve

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

Solve

3^{x} = 3 : x = 1

x = \frac{ln ( 6 )}{ln ( 3 )}, x = 1

5^{4 - x} = 0

8 (4^{6 - 2 x}) + 13 = 41

7^{x} = 2 \cdot 5^{x}

6^{x} = \frac{1}{6}

200 = \frac{500}{1 + e ^{3 - t}}

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