What is the answer to 3^{2x+1}-7*3^x+4=0 ?

The answer to 3^{2x+1}-7*3^x+4=0 is x=(ln(4/3))/(ln(3)),x=0

Solve for x: 3^{2x+1}-7*3^x+4=0

The solution is x=(ln(4/3))/(ln(3)),x=0

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3^{2 x + 1} - 7 \cdot 3^{x} + 4 = 0

x = \frac{ln ( \frac{4}{3} )}{ln ( 3 )}, x = 0

Decimal Notation

x = 0.26185 \dots, x = 0

Solution steps

3^{2 x + 1} - 7 \cdot 3^{x} + 4 = 0

Apply exponent rules

(3^{x})^{2} \cdot 3^{1} - 7 \cdot 3^{x} + 4 = 0

Rewrite the equation with

3^{x} = u

(u)^{2} \cdot 3^{1} - 7 u + 4 = 0

Solve

u^{2} \cdot 3^{1} - 7 u + 4 = 0 : u = \frac{4}{3}, u = 1

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

Substitute back

u = 3^{x},

solve for

x

Solve

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

Solve

3^{x} = 1 : x = 0

x = \frac{ln ( \frac{4}{3} )}{ln ( 3 )}, x = 0

e^{x} - e^{- x} = \frac{3}{2}

x, 3 y = - 5^{x} + 7

3^{5 x - 4} = 21

2^{4 x - 3} = 18

What is the answer to 3^{2x+1}-7*3^x+4=0 ?
The answer to 3^{2x+1}-7*3^x+4=0 is x=(ln(4/3))/(ln(3)),x=0
Solve for x: 3^{2x+1}-7*3^x+4=0
The solution is x=(ln(4/3))/(ln(3)),x=0