What are the solutions to the equation 19=32(1-b)^{26} ?

The solutions to the equation 19=32(1-b)^{26} are b=-\sqrt[26]{19/32}+1,b=\sqrt[26]{19/32}+1

Find the zeros of 19=32(1-b)^{26}

The zeros of 19=32(1-b)^{26} are b=-\sqrt[26]{19/32}+1,b=\sqrt[26]{19/32}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

19 = 32 (1 - b)^{26}

Decimal Notation

b = 0.01985 \dots, b = 1.98014 \dots

Solution steps

19 = 32 (1 - b)^{26}

Switch sides

32 (1 - b)^{26} = 19

Divide both sides by

32

(1 - b)^{26} = \frac{19}{32}

For

(g (x))^{n} = f (a)

, n is even, the solutions are

Solve

The solutions are

x^{3} + x = - 2

x^{3} + x = 10

x^{5} - 3 x^{3} - 4 x^{2} + 2 x + 4 = 0

n^{3} = 60

- x^{3} + 3 x^{2} + 6 x - 8 = 0

What are the solutions to the equation 19=32(1-b)^{26} ?
The solutions to the equation 19=32(1-b)^{26} are b=-\sqrt[26]{19/32}+1,b=\sqrt[26]{19/32}+1
Find the zeros of 19=32(1-b)^{26}
The zeros of 19=32(1-b)^{26} are b=-\sqrt[26]{19/32}+1,b=\sqrt[26]{19/32}+1