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

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Final answer to the problem

The equation has no solutions.

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Apply the formula: $a\log_{b}\left(x\right)$$=\log_{b}\left(x^a\right)$, where $a=2$, $b=10$ and $x=x+2$

Learn how to solve factor by difference of squares problems step by step online.

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

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Learn how to solve factor by difference of squares problems step by step online. Solve the logarithmic equation log(x+-1)+log(x+1)=2log(x+2). Apply the formula: a\log_{b}\left(x\right)=\log_{b}\left(x^a\right), where a=2, b=10 and x=x+2. The sum of two logarithms of the same base is equal to the logarithm of the product of the arguments. Solve the product of difference of squares \left(x-1\right)\left(x+1\right). For two logarithms of the same base to be equal, their arguments must be equal. In other words, if \log(a)=\log(b) then a must equal b.

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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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