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Solve the logarithmic equation $\log \left(x-1\right)+\log \left(x+1\right)=2\log \left(x+2\right)$

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Solution

Videos

Tutorial - Solve by factoring when a perfect square 0=x^2+6x+9

https://www.youtube.com/watch?v=F7O02LqPQp0

Algebra 2 - Solving a factored quadratic equation using the zero product property, -x(3x + 5) = 0

https://www.youtube.com/watch?v=3on6rsKoK_k

Algebra 2 - Solving by factoring and using the square root method x^2 + 16x + 64 = 15

https://www.youtube.com/watch?v=WGuLJdN2Tuo

Algebra 2 - using the square root method to solve a quadratic equation 4x^2 - 4x + 1 = 9

https://www.youtube.com/watch?v=90FQ2oLoEEQ

Algebra 2 - How to find the solutions of a quadratic using the quadratic formula, 4x^2 + x = 3

https://www.youtube.com/watch?v=ZQu6QjBgXS8

Tutorial - How to solve by factoing using a perfect square trinomial y=20x+x^2+100

https://www.youtube.com/watch?v=VV8EOcSVZX4

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Similar Problems Solved

$\log\left(x-1\right)+\log\left(x+1\right)=2\log\left(x+2\right)$

$\log x+\log\left(x-3\right)=1$

$2log\left(x\right)-log\left(x+6\right)=0$

$\log_x\left(64\right)=3$

$\log_2\left(x-2\right)+\log_2\left(x+1\right)=2$

$2\cdot log\left(x\right)-1\cdot log\left(x+6\right)=0$

$log\left(x+1\right)=log\left(x-1\right)+3$

Main Topic: Factor by Difference of Squares

The difference of two squares is a squared number subtracted from another squared number. Every difference of squares may be factored according to the identity a^2-b^2=(a+b)(a-b) in elementary algebra.

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