What is the answer to 5^{2x}-5^{x+1}+4=0 ?

The answer to 5^{2x}-5^{x+1}+4=0 is x=(2ln(2))/(ln(5)),x=0

Solve for x: 5^{2x}-5^{x+1}+4=0

The solution is x=(2ln(2))/(ln(5)),x=0

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5^{2 x} - 5^{x + 1} + 4 = 0

x = \frac{2 ln ( 2 )}{ln ( 5 )}, x = 0

Decimal Notation

x = 0.86135 \dots, x = 0

Solution steps

5^{2 x} - 5^{x + 1} + 4 = 0

Apply exponent rules

(5^{x})^{2} - 5^{x} \cdot 5^{1} + 4 = 0

Rewrite the equation with

5^{x} = u

(u)^{2} - u \cdot 5^{1} + 4 = 0

Solve

u^{2} - u \cdot 5^{1} + 4 = 0 : u = 4, u = 1

u = 4, u = 1

Substitute back

u = 5^{x},

solve for

x

Solve

5^{x} = 4 : x = \frac{2 ln ( 2 )}{ln ( 5 )}

Solve

5^{x} = 1 : x = 0

x = \frac{2 ln ( 2 )}{ln ( 5 )}, x = 0

3^{x^{2}} \cdot 3^{5 x} = 729

2 7^{\frac{2}{5}} = 3^{x}

4^{x^{3}} = 2^{54}

3^{x + 1} + 4 = 8

e^{2} x - 4 e^{x} - 5 = 0

What is the answer to 5^{2x}-5^{x+1}+4=0 ?
The answer to 5^{2x}-5^{x+1}+4=0 is x=(2ln(2))/(ln(5)),x=0
Solve for x: 5^{2x}-5^{x+1}+4=0
The solution is x=(2ln(2))/(ln(5)),x=0