What is the answer to 7^{2x}+7^{x+1}-18=0 ?

The answer to 7^{2x}+7^{x+1}-18=0 is x=(ln(2))/(ln(7))

Solve for x: 7^{2x}+7^{x+1}-18=0

The solution is x=(ln(2))/(ln(7))

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7^{2 x} + 7^{x + 1} - 18 = 0

x = \frac{ln ( 2 )}{ln ( 7 )}

Decimal Notation

x = 0.35620 \dots

Solution steps

7^{2 x} + 7^{x + 1} - 18 = 0

Apply exponent rules

(7^{x})^{2} + 7^{x} \cdot 7^{1} - 18 = 0

Rewrite the equation with

7^{x} = u

(u)^{2} + u \cdot 7^{1} - 18 = 0

Solve

u^{2} + u \cdot 7^{1} - 18 = 0 : u = 2, u = - 9

u = 2, u = - 9

Substitute back

u = 7^{x},

solve for

x

Solve

7^{x} = 2 : x = \frac{ln ( 2 )}{ln ( 7 )}

Solve

7^{x} = - 9 :

No Solution for

x \in R

x = \frac{ln ( 2 )}{ln ( 7 )}

e^{- 2 x} + 6 = 5 e^{- x}

5 e^{- 2 t} = 6

m, 5^{n} + 5^{n} + 5^{n} + 5^{n} + 5^{n} = 5^{m}

312 5^{x + 4} = 2 5^{2 x - 2}

5 e^{0.01 t} = 2 e^{0.02 t}

What is the answer to 7^{2x}+7^{x+1}-18=0 ?
The answer to 7^{2x}+7^{x+1}-18=0 is x=(ln(2))/(ln(7))
Solve for x: 7^{2x}+7^{x+1}-18=0
The solution is x=(ln(2))/(ln(7))