log₄(10) - log₄(x + 3) = log₄(x)

Answer

Explanation:

Step1: Apply log subtraction rule

Using the property (\log_a M - \log_a N=\log_a\frac{M}{N}), we rewrite the left - hand side of the equation (\log_4(10)-\log_4(x + 3)=\log_4(x)) as (\log_4\left(\frac{10}{x + 3}\right)=\log_4(x))

Step2: Use one - to - one property of logs

If (\log_a b=\log_a c), then (b = c) (for (a>0,a\neq1,b>0,c>0)). So we have the equation (\frac{10}{x + 3}=x)

Step3: Solve the rational equation

Multiply both sides of the equation (\frac{10}{x + 3}=x) by (x + 3) (assuming (x+3\neq0) i.e., (x\neq - 3)) to get (10=x(x + 3)) Expand the right - hand side: (10=x^{2}+3x) Rearrange the equation to the standard quadratic form: (x^{2}+3x - 10 = 0)

Step4: Factor the quadratic equation

Factor (x^{2}+3x - 10). We need two numbers that multiply to (- 10) and add up to (3). The numbers are (5) and (-2). So (x^{2}+3x - 10=(x + 5)(x - 2)=0)

Step5: Find the roots of the quadratic equation

Set each factor equal to zero:

(x+5 = 0) gives (x=-5)
(x - 2=0) gives (x = 2)

Step6: Check the domain

For the original logarithmic equation (\log_4(10)-\log_4(x + 3)=\log_4(x)), we need:

(x+3>0) (so (x>-3))
(x>0)

For (x=-5): (x=-5) does not satisfy (x>0) and (x>-3), so we reject (x = - 5) For (x = 2): (x+3=2 + 3=5>0) and (x = 2>0), so (x = 2) is a valid solution.

Answer:

(x = 2)