What is the answer to log_{2-x}(2x^2-5x+2)=2 ?

The answer to log_{2-x}(2x^2-5x+2)=2 is x=-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

log_{2-x}(2x^2-5x+2)=2

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

x = - 1

Solution steps

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

Apply log rules

\frac{ln ( 2 x ^{2} - 5 x + 2 )}{ln ( 2 - x )} = 2

Multiply both sides by

ln (2 - x)

\frac{ln ( 2 x ^{2} - 5 x + 2 )}{ln ( 2 - x )} ln (2 - x) = 2 ln (2 - x)

Simplify

ln (2 x^{2} - 5 x + 2) = 2 ln (2 - x)

Apply log rules

2 x^{2} - 5 x + 2 = (2 - x)^{2}

Solve

2 x^{2} - 5 x + 2 = (2 - x)^{2} : x = 2, x = - 1

x = 2, x = - 1

Verify Solutions:

x = 2

False

, x = - 1

True

The solution is

x = - 1

4 = - lo g_{10} (x)

lo g_{8} (4 x + 4) = 2

x, y^{2} + ln (x + y^{2}) = x + 3

12 x^{2} ln (x) + 7 x^{2} = 0

lo g_{10} (3 x + 1) = 2

What is the answer to log_{2-x}(2x^2-5x+2)=2 ?
The answer to log_{2-x}(2x^2-5x+2)=2 is x=-1
Solve for x: log_{2-x}(2x^2-5x+2)=2
The solution is x=-1