How do you solve #log_[2] (x+20) - log_[2] (x+2) = log_[2] x#?

First of all, let's establish the domain. As any logarithmic expression can only have positive arguments, the following conditions must hold:

• In order for the first logarithmic expression to be defined, #x + 20 > 0 " "<=>" " x > -20# must hold.
• For the second logarithmic expression, #x + 2 > 0 <=> x > -2# must hold.
• For the third logarithmus, #x > 0# must hold.

As #x > 0# is the strictest condition of the three (all three conditions would be satisfied if this one is satisfied).

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Now, let's solve the equation.

First of all, we can use the logarithmic law

#log_a (m) - log_a(n) = log_a (m/n)#

In this case, we get:

#log_2(x+20) - log_2(x+2) = log_2(x)#

#<=> log_2 ((x+20)/(x+2)) = log_2(x)#

Now that on both sides we have just one logarithmic expression, we can use #log x = log y# if and only if #x = y# and drop the logarithms:

#<=> (x+20)/(x+2) = x#

... multiply both sides with #(x +2)#...

#<=> x + 20 = x(x+2)#

#<=> x^2 + x -20 = 0#

The solutions for this quadratic equation are #x = -5# or #x = 4#.

However, as we have established that #x > 0# must hold in order for our logarithmic expressions to be defined, #x = -5# can't be a solution.

Thus, our only solution is #x = 4#.

Try this: