`log_4(-x)+log_4(x+10)=2` Solve the equation. Check for extraneous solutions.

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marizi eNotes educator| Certified Educator

To evaluate the given equation `log_4(-x)+log_4(x+10)=2` , we may apply the logarithm property: `log_b(x)+log_b(y)=log_b(x*y)` .

`log_4(-x)+log_4(x+10)=2`

`log_4((-x)*(x+10))=2`

`log_4(-x^2-10x)=2`

To get rid of the "log" function, we may apply the logarithm property: `b^(log_b(x))=x.`

Raise both sides by base of `4` .

`4^(log_4(-x^2-10x))=4^2`

`-x^2-10x=16`

Add `x^2` and `10x` on both sides of the equation to simplify in standard form: `ax^2+bx+c= 0.`

`-x^2-10x+x^2+10x=16+x^2+10x`

`0=16+x^2+10x orx^2+10x+16=0.`

Apply factoring on the trinomial.

`(x+2)(x+8)=0`

Apply zero-factor property to solve for x by equating each factor to `0` .

`x+2=0`

`x+2-2=0-2`

`x=-2`

and

`x+8=0`

` x+8-8=0-8 `

`x=-8`

Checking: Plug-in each `x ` on `log_4(-x)+log_4(x+10)=2` .

Let `x=-2` on ` log_4(-x)+log_4(x+10)=2` .

`log_4(-(-2))+log_4(-2+10)=?2`

`log_4(2)+log_4(8)=?2`

`log_4(2*8)=?2`

`log_4(16)=?2`

`log_4(4^2)=?2`

`2log_4(4)=?2`

`2*1=?2`

`2=2`        TRUE

Let `x=-8` on `log_4(-x)+log_4(x+10)=2.`

`log_4(-(-8))+log_4(-8+10)=?2`

`log_4(8)+log_4(2)=?2`

`log_4(8*2)=?2`

`log_4(16)=?2`

`2=2 `        TRUE

Therefore, there are no extraneous solutions.

Both solved x-values: `x=-2` and `x=-8` are real solution of the equation `log_4(-x)+log_4(x+10)=2` .

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