How do you solve #log_[2] (x+20) - log_[2] (x+2) = log_[2] x#?
2 Answers
Explanation:
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
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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
#<=> (x+20)/(x+2) = x#
... multiply both sides with
#<=> x + 20 = x(x+2)#
#<=> x^2 + x -20 = 0#
The solutions for this quadratic equation are
However, as we have established that
Thus, our only solution is
I found
Explanation:
Try this:
