Question

Which of the following shows the extraneous solution to the logarithmic equation? log Subscript 4 Baseline (x) + log Subscript 4 Baseline (x minus 3) = log Subscript 4 Baseline (negative 7 x + 21)

Answer 1

C

Explanation:

Edge 2021

Answer 2

Given:

The given equation is
`\log _(4)(x)+\log _(4)(x-3)=\log _(4)(-7 x+21)`

We need to determine the extraneous solution of the equation.

Solving the equation:

To determine the extraneous solution, we shall first solve the given equation.

Applying the log rule
`\log _(c)(a)+\log _(c)(b)=\log _(c)(a b)`, we get;

`\log _(4)(x(x-3))=\log _(4)(-7 x+21)`

Again applying the log rule, if
`\log _(b)(f(x))=\log _(b)(g(x))` then
`f(x)=g(x)`

Thus, we have;

`x(x-3)=-7 x+21`

Simplifying the equation, we get;

`x^2-3x=-7 x+21`

`x^2+4x=21`

`x^2+4x-21=0`

Factoring the equation, we get;

`(x-3)(x+7)=0`

Thus, the solutions are
`x=3, x=-7`

Extraneous solutions:

The extraneous solutions are the solutions that does not work in the original equation.

Now, to determine the extraneous solution, let us substitute x = 3 and x = -7 in the original equation.

Thus, we get;

`\log _(4)(3)+\log _(4)(3-3)=\log _(4)(-7 \cdot 3+21)`

`\log _(4)(3)+\log _(4)(0)=\log _(4)(0)`

Since, we know that
`\log _(a)(0)` is undefined.

Thus, we get;

Undefined = Undefined

This is false.

Thus, the solution x = 3 does not work in the original equation.

Hence, x = 3 is an extraneous solution.

Similarly, substituting x = -7, in the original equation. Thus, we get;

`\log _(4)(-7)+\log _(4)(-7-3)=\log _(4)(-7(-7)+21)`

`\log _(4)(-7)+\log _(4)(-10)=\log _(4)(49+21)`

`\log _(4)(-7)+\log _(4)(-10)=\log _(4)(70)`

Simplifying, we get;

Undefined =
`\log _(4)(70)`

Undefined = 3.06

This is false.

Thus, the solution x = -7 does not work in the original solution.

Hence, x = -7 is an extraneous solution.

Therefore, the extraneous solutions are x = 3 and x = -7