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

The answer to log_{-x-1}(2x^2-x+1)=2 is No Solution for x\in\mathbb{R}

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

The solution is No Solution for x\in\mathbb{R}

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

No Solution for x \in R

Solution steps

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

Apply log rules

\frac{ln ( 2 x ^{2} - x + 1 )}{ln ( - x - 1 )} = 2

Multiply both sides by

ln (- x - 1)

\frac{ln ( 2 x ^{2} - x + 1 )}{ln ( - x - 1 )} ln (- x - 1) = 2 ln (- x - 1)

Simplify

ln (2 x^{2} - x + 1) = 2 ln (- x - 1)

Apply log rules

2 x^{2} - x + 1 = (- x - 1)^{2}

Solve

2 x^{2} - x + 1 = (- x - 1)^{2} : x = 3, x = 0

x = 3, x = 0

Verify Solutions:

x = 3

False

, x = 0

False

The solution is

No Solution for x \in R

lo g_{9} (2 x - 5) = 0.5

lo g_{9} (27 x) + lo g_{x} (81) = 6

lo g_{10} (x) 27 = - 3

ln (x - 6) + ln (x + 1) = ln (x - 15)

ln (x) + ln (x + 4) = ln (12)

What is the answer to log_{-x-1}(2x^2-x+1)=2 ?
The answer to log_{-x-1}(2x^2-x+1)=2 is No Solution for x\in\mathbb{R}
Solve for x: log_{-x-1}(2x^2-x+1)=2
The solution is No Solution for x\in\mathbb{R}