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Solve the logarithmic equation $\ln\left(x^3-4\right)-\ln\left(x^2-2x\right)=\ln\left(x+2\right)$

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Solution

Videos

Applying the difference of two squares with fractions, (1/4)x^2 - (1/4)

https://www.youtube.com/watch?v=iCJqYiruM98

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

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

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

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

Algebra 2 - Learn how to complete the square to solve a quadratic, x^2 - 14x + 19 = 0

https://www.youtube.com/watch?v=xRlis69-HA0

How to factor a quadratic equation by using a perfect square

https://www.youtube.com/watch?v=wb4T0j_C8bU

Applying equality property of logarithms to solve by factoring, log4 (x^2 -4)=log4 (-x+2)

https://www.youtube.com/watch?v=z7AHkGgPHxQ

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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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