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

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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)` .




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` .



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


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

Apply factoring on the trinomial.


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






` x+8-8=0-8 `


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` .








`2=2`        TRUE

Let `x=-8` on `log_4(-x)+log_4(x+10)=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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