What is the answer to 3^2x+1-10(3^x)+3=0 ?

The answer to 3^2x+1-10(3^x)+3=0 is No Solution for x\in\mathbb{R}

Solve for x: 3^2x+1-10(3^x)+3=0

The solution is No Solution for x\in\mathbb{R}

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3^{2} x + 1 - 10 (3^{x}) + 3 = 0

No Solution for x \in R

Solution steps

3^{2} x + 1 - 10 (3^{x}) + 3 = 0

Prepare

3^{2} x + 1 - 10 (3^{x}) + 3 = 0

for Lambert form:

- 10 = - e^{- l n (3) x} (9 x + 4)

Rewrite the equation with

(- x - \frac{4}{9}) ln (3) = u

and

x = - \frac{9 u + 4 ln ( 3 )}{9 ln ( 3 )}

- 10 = - e^{- l n (3) (- \frac{9 u + 4 l n ( 3 )}{9 l n ( 3 )})} (9 (- \frac{9 u + 4 ln ( 3 )}{9 ln ( 3 )}) + 4)

Simplify

- 10 = - e^{\frac{9 u + 4 l n ( 3 )}{9}} (- \frac{9 u + 4 ln ( 3 )}{ln ( 3 )} + 4)

Rewrite

- 10 = - e^{\frac{9 u + 4 l n ( 3 )}{9}} (- \frac{9 u + 4 ln ( 3 )}{ln ( 3 )} + 4)

in Lambert form:

\frac{e ^{\frac{9 u + 4 l n ( 3 )}{9}} u}{e ^{\frac{4 l n ( 3 )}{9}}} = - \frac{10 ln ( 3 )}{9 e ^{\frac{4 l n ( 3 )}{9}}}

Solve

\frac{e ^{\frac{9 u + 4 l n ( 3 )}{9}} u}{e ^{\frac{4 l n ( 3 )}{9}}} = - \frac{10 ln ( 3 )}{9 e ^{\frac{4 l n ( 3 )}{9}}} :

No Solution for

u \in R

No Solution for x \in R

e^{5 x} = e^{15}

5^{2 x} - 5^{x} - 20 = 0

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

6^{7 x} = 8

e^{x} + 20 e^{- x} = 9

What is the answer to 3^2x+1-10(3^x)+3=0 ?
The answer to 3^2x+1-10(3^x)+3=0 is No Solution for x\in\mathbb{R}
Solve for x: 3^2x+1-10(3^x)+3=0
The solution is No Solution for x\in\mathbb{R}