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

The answer to 2^{x+1}=3^{2-2x} is x=(2ln(3)-ln(2))/(ln(18))

Solve for x: 2^{x+1}=3^{2-2x}

The solution is x=(2ln(3)-ln(2))/(ln(18))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

x = \frac{2 ln ( 3 ) - ln ( 2 )}{ln ( 18 )}

Decimal Notation

x = 0.52037 \dots

Solution steps

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

Apply exponent rules

(x + 1) ln (2) = (2 - 2 x) ln (3)

Solve

(x + 1) ln (2) = (2 - 2 x) ln (3) : x = \frac{2 ln ( 3 ) - ln ( 2 )}{ln ( 18 )}

x = \frac{2 ln ( 3 ) - ln ( 2 )}{ln ( 18 )}

x, (x y)^{x} = e

0. 5^{x} = 0.15

5^{2 x - 1} = 0.2

1 0^{2 x + 1} = 115

100 = 75 e^{0.05 x}

What is the answer to 2^{x+1}=3^{2-2x} ?
The answer to 2^{x+1}=3^{2-2x} is x=(2ln(3)-ln(2))/(ln(18))
Solve for x: 2^{x+1}=3^{2-2x}
The solution is x=(2ln(3)-ln(2))/(ln(18))