What is the answer to 1+log_{5}(x^2+1)=log_{5}(x^2+4x+1) ?

The answer to 1+log_{5}(x^2+1)=log_{5}(x^2+4x+1) is No Solution for x\in\mathbb{R}

Solve for x: 1+log_{5}(x^2+1)=log_{5}(x^2+4x+1)

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

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1 + lo g_{5} (x^{2} + 1) = lo g_{5} (x^{2} + 4 x + 1)

No Solution for x \in R

Solution steps

1 + lo g_{5} (x^{2} + 1) = lo g_{5} (x^{2} + 4 x + 1)

Apply log rules

5 (x^{2} + 1) = x^{2} + 4 x + 1

Solve

5 (x^{2} + 1) = x^{2} + 4 x + 1 :

No Solution for

x \in R

No Solution for x \in R

x, 2 lo g_{4} (x) - lo g_{2} (y) = 3

ln (2 x - 2) = 0

ln (2 x - 2) = 2

lo g_{2} (x) + lo g_{2} (x^{2}) + lo g_{3} (x^{3}) = 6

lo g_{10} (x) (25) = lo g_{5} (x)

What is the answer to 1+log_{5}(x^2+1)=log_{5}(x^2+4x+1) ?
The answer to 1+log_{5}(x^2+1)=log_{5}(x^2+4x+1) is No Solution for x\in\mathbb{R}
Solve for x: 1+log_{5}(x^2+1)=log_{5}(x^2+4x+1)
The solution is No Solution for x\in\mathbb{R}