Need help with this domain question.

[s garg]

15. Find all the values of the argument a with which the equation

tan (2π (x +1/2)) ⋅ log(4x + a + 3) = log(4x + a + 3)

has the only root in the interval [−1/2; 0] .

 

[pka]

15. Find all the values of the argument a with which the equation
tan (2π (x +1/2)) ⋅ log(4x + a + 3) = log(4x + a + 3)
has the only root in the interval [−1/2; 0] .

To s garg, have you posted a correct copy of you question?
If so, it implies that \(\displaystyle \tan\left(2\pi(x+\tfrac{1}{2})\right)=1\) Do you see why?
LOOK HERE

 

[s garg]

To s garg, have you posted a correct copy of you question?
If so, it implies that \(\displaystyle \tan\left(2\pi(x+\tfrac{1}{2})\right)=1\) Do you see why?
LOOK HERE

yes sir that is why i asked this question. and yes it is the correct copy of the problem. The question asks for the values of a but here as you said the whole term with " a " is getting eliminated. so i am confused. Can we really eliminate the log function like that?

 

[MarkFL]

15. Find all the values of the argument a with which the equation

tan (2π (x +1/2)) ⋅ log(4x + a + 3) = log(4x + a + 3)

has the only root in the interval [−1/2; 0] .

I would arrange and factor as follows:

[MATH]\log(4x+a+3)\left(\tan\left(2\pi\left(x+\frac{1}{2}\right)\right)-1\right)=0[/MATH]
And so, we have two cases to consider, but only 1 involves the parameter \(a\):

[MATH]\log(4x+a+3)=0\implies x=-\frac{a+2}{4}[/MATH]
[MATH]-\frac{1}{2}\le x\le0[/MATH]
[MATH]-\frac{1}{2}\le -\frac{a+2}{4}\le0[/MATH]
Can you finish?

 

[s garg]

I would arrange and factor as follows:

[MATH]\log(4x+a+3)\left(\tan\left(2\pi\left(x+\frac{1}{2}\right)\right)-1\right)=0[/MATH]
And so, we have two cases to consider, but only 1 involves the parameter \(a\):

[MATH]\log(4x+a+3)=0\implies x=-\frac{a+2}{4}[/MATH]
[MATH]-\frac{1}{2}\le x\le0[/MATH]
[MATH]-\frac{1}{2}\le -\frac{a+2}{4}\le0[/MATH]
Can you finish?

Thank you sir! and yes i can finish now i never thought of rearranging it and was stuck just by eliminating the log terms. i see now thank you alot!