What is the answer to 2^{1-6x}=7^{3x+4} ?

The answer to 2^{1-6x}=7^{3x+4} is x=-(4ln(7)-ln(2))/(3ln(28))

Solve for x: 2^{1-6x}=7^{3x+4}

The solution is x=-(4ln(7)-ln(2))/(3ln(28))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2^{1 - 6 x} = 7^{3 x + 4}

x = - \frac{4 ln ( 7 ) - ln ( 2 )}{3 ln ( 28 )}

Decimal Notation

x = - 0.70928 \dots

Solution steps

2^{1 - 6 x} = 7^{3 x + 4}

Apply exponent rules

(1 - 6 x) ln (2) = (3 x + 4) ln (7)

Solve

(1 - 6 x) ln (2) = (3 x + 4) ln (7) : x = - \frac{4 ln ( 7 ) - ln ( 2 )}{3 ln ( 28 )}

x = - \frac{4 ln ( 7 ) - ln ( 2 )}{3 ln ( 28 )}

y, 9^{x + 3} - 2 7^{2 y} = - 3535662

e^{x} = 1 + x

(2^{2})^{n} = 16

8^{6 y + 4} = 64

3^{x} + 3^{x + 3} = 756

What is the answer to 2^{1-6x}=7^{3x+4} ?
The answer to 2^{1-6x}=7^{3x+4} is x=-(4ln(7)-ln(2))/(3ln(28))
Solve for x: 2^{1-6x}=7^{3x+4}
The solution is x=-(4ln(7)-ln(2))/(3ln(28))