What is the answer to 5^{2x}+5^{x+1}-14=0 ?

The answer to 5^{2x}+5^{x+1}-14=0 is x=(ln(2))/(ln(5))

Solve for x: 5^{2x}+5^{x+1}-14=0

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

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5^{2 x} + 5^{x + 1} - 14 = 0

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

Solution steps

5^{2 x} + 5^{x + 1} - 14 = 0

Apply exponent rules

(5^{x})^{2} + 5^{x} \cdot 5^{1} - 14 = 0

Rewrite the equation with

5^{x} = u

(u)^{2} + u \cdot 5^{1} - 14 = 0

Solve

u^{2} + u \cdot 5^{1} - 14 = 0 : u = 2, u = - 7

u = 2, u = - 7

Substitute back

u = 5^{x},

solve for

x

Solve

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

Solve

5^{x} = - 7 :

No Solution for

x \in R

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

2^{x + 3} - 15 = 2^{1 - x}

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

36 e^{5 - x} - 6^{2 x} = 0

1 0^{x} \cdot 1 0^{x + 2} = 1000

e^{5 x} = 17

What is the answer to 5^{2x}+5^{x+1}-14=0 ?
The answer to 5^{2x}+5^{x+1}-14=0 is x=(ln(2))/(ln(5))
Solve for x: 5^{2x}+5^{x+1}-14=0
The solution is x=(ln(2))/(ln(5))