What is the answer to 12=-10log_{10}((1-x)^2) ?

The answer to 12=-10log_{10}((1-x)^2) is x=-sqrt(1/(10\sqrt[5]{10))}+1,x=sqrt(1/(10\sqrt[5]{10))}+1

Solve for x: 12=-10log_{10}((1-x)^2)

The solution is x=-sqrt(1/(10\sqrt[5]{10))}+1,x=sqrt(1/(10\sqrt[5]{10))}+1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

12 = - 10 lo g_{10} ((1 - x)^{2})

Decimal Notation

x = 0.74881 \dots, x = 1.25118 \dots

Solution steps

12 = - 10 lo g_{10} ((1 - x)^{2})

Switch sides

- 10 lo g_{10} ((1 - x)^{2}) = 12

Divide both sides by

- 10

lo g_{10} ((1 - x)^{2}) = - \frac{6}{5}

Apply log rules

Solve

Verify Solutions:

True

The solutions are

ln (γ) = ln (4 + e^{2^{β}}) + ω

lo g_{7} (\frac{1}{x}) = - 3

lo g_{2} (x + 3) + lo g_{2} (x - 5) = 0

lo g_{5} (x + 6) = - lo g_{5} (x + 2) + 1

lo g_{4} (x + 11) = 2

What is the answer to 12=-10log_{10}((1-x)^2) ?
The answer to 12=-10log_{10}((1-x)^2) is x=-sqrt(1/(10\sqrt[5]{10))}+1,x=sqrt(1/(10\sqrt[5]{10))}+1
Solve for x: 12=-10log_{10}((1-x)^2)
The solution is x=-sqrt(1/(10\sqrt[5]{10))}+1,x=sqrt(1/(10\sqrt[5]{10))}+1