problem finding a domain

Hermitage

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Hello.

i can't get to solve a domain problem.

0<sqrt(1-log(5x-2))<1

How am I to find the x solution set?

The more detailed answer the better, because i just cant wrap my head around it. Thankyou!
 
i can't get to solve a domain problem.

0<sqrt(1-log(5x-2))<1
This looks like a "solving a radical inequality" problem. How is this actually a "find the domain of the function" problem?

How am I to find the x solution set?

The more detailed answer the better....
You already have loads of worked examples in your book, in your class notes, and in the various online lessons you've taken the time to study. So the worked solution to one more exercise isn't likely to make much of a difference. Instead, let's try to help you work through this solution!

First, what did you see when you graphed the associated function, \(\displaystyle f(x)\, =\, y\, =\, \sqrt{1\, -\, \log(5x\, -\, 2)\, }\)?

You know that any square root has to evaluate to a non-negative value, so the "greater than zero" restriction only means "not equal to zero". What would the value of the insides of the square root need to be, in order for the square root to have a value of zero? Set the insides equal to this, and solve. This is one endpoint of the solution; you'll need to remember that this endpoint is not included in the solution.

Please reply with your efforts on the above. Then we'll see where we need to go from there. Thank you! ;)
 
Thank you for your time!

I took a bit of a shortcut with this exercise, and i have most on it done. Heck i even reverse engineered the answer to specifically understand that logarithm part but dont get it still. I mean i know what log is what it does, how its calculated but i dont know how must I restrict the 5x-2 to meet the restrictions... What i figured, though havent done yet is to 0<log(5x-2)<1

Ok but let me start from the beginning:

The problem:

1/(e^sqrt(4x+3) * sqrt(1-log(5x-2))
MSP1154203fa7fdhdi3675g00003cdghe5d82ah51ie

and yes, i need to find the domain of this thing (all the x values right?)

So my observations:
1) lower part can not be 0
2) from first sqrt we get> X>=-3/4
3) second is the funny one. i can shove away the log restriction because i basically have to solve 0<log(5x-2)<1

3.1) The heck im supposed to do now, because the answer is: X=]-0.75, 1.6] how should i make x<=1.6 from 5x-2 ??
 
Thank you for your time!

I took a bit of a shortcut with this exercise, and i have most on it done. Heck i even reverse engineered the answer to specifically understand that logarithm part but dont get it still. I mean i know what log is what it does, how its calculated but i dont know how must I restrict the 5x-2 to meet the restrictions... What i figured, though havent done yet is to 0<log(5x-2)<1

Ok but let me start from the beginning:

The problem:

1/(e^sqrt(4x+3) * sqrt(1-log(5x-2))
MSP1154203fa7fdhdi3675g00003cdghe5d82ah51ie

and yes, i need to find the domain of this thing (all the x values right?)

So my observations:
1) lower part can not be 0
2) from first sqrt we get> X>=-3/4
3) second is the funny one. i can shove away the log restriction because i basically have to solve 0<log(5x-2)<1

3.1) The heck im supposed to do now, because the answer is: X=]-0.75, 1.6] how should i make x<=1.6 from 5x-2 ??


There is something wrong with the interpretation of the problem!

The argument of the log function must be positive in real domain. So you cannot shove away the log restriction.

5x -2 > 0 → x > 0.4 (not x ≥ -0.75)
 
There is something wrong with the interpretation of the problem!

The argument of the log function must be positive in real domain. So you cannot shove away the log restriction.

5x -2 > 0 → x > 0.4 (not x ≥ -0.75)


Dear friend you completely forget about the first square root that is above e

I will post my problem again and explain again my thoughts on it.
View attachment mata.bmp

What i need to find?
I need to find all x values that allow this expression to be evaluated. That is finding DOMAIN right?

What i do?

Well from this part e^sqrt(4x+3) i get that x must be always x ≥ -0.75.

The problem I am facing:

I understand that next x values restrictions must come from here sqrt(1-log(5x-2))
1) as all the expression is under sqrt sign it means that 1-log(5x-2)>=0
2) that means that log(5x-2)<= 1 (that is why i said i can shove the log restriction away. Its because i need log to evaluate between 0 and 1 anyway. Strictly speaking I ofcourse must take the log restriction in account, i can assure you I do, but on the other hand that is not the problem i have, so i wanted to say that you do not have to concern yourself with it)

So all in all, WHAT ARE THE x values that make log(5x-2) evaluate between 0 and 1. How can i express that?

My thought on that are: expression in the bracket cant be negative. From here i now get 5x>2 therefore x>2/5 but that does not help me much, as i need the x value that will make log(5x-2) evaluate to 1. What i need to do to get that x value that will make log evaluate to 1??? How must i write it down??
 
Dear friend you completely forget about the first square root that is above e

I will post my problem again and explain again my thoughts on it.
View attachment 4306

What i need to find?
I need to find all x values that allow this expression to be evaluated. That is finding DOMAIN right?

What i do?

Well from this part e^sqrt(4x+3) i get that x must be always x ≥ -0.75.

The problem I am facing:

I understand that next x values restrictions must come from here sqrt(1-log(5x-2))
1) as all the expression is under sqrt sign it means that 1-log(5x-2)>=0
2) that means that log(5x-2)<= 1 (that is why i said i can shove the log restriction away. Its because i need log to evaluate between 0 and 1 anyway. Strictly speaking I ofcourse must take the log restriction in account, i can assure you I do, but on the other hand that is not the problem i have, so i wanted to say that you do not have to concern yourself with it)

So all in all, WHAT ARE THE x values that make log(5x-2) evaluate between 0 and 1. How can i express that?

My thought on that are: expression in the bracket cant be negative. From here i now get 5x>2 therefore x>2/5 but that does not help me much, as i need the x value that will make log(5x-2) evaluate to 1. What i need to do to get that x value that will make log evaluate to 1??? How must i write it down??

What is the value of log(5x-2) at x = 0? Do you see know that the minimum value of x is NOT -0.75 (because 0>-0.75).

Anyway regarding your other question:

1 - log(5x-2) > 0

Do you know the definition of Log? In particular:

Loga(b) = c → b = ac

That should help you get the upper limit.
 
What is the value of log(5x-2) at x = 0? Do you see know that the minimum value of x is NOT -0.75 (because 0>-0.75).

Anyway regarding your other question:

1 - log(5x-2) > 0

Do you know the definition of Log? In particular:

Loga(b) = c → b = ac

That should help you get the upper limit.

Good point! The answer in the book is wrong for the -0,75 part.

On the log part:
i take the log base = 10. Since nothing else is written there i assume its 10.
so
10^0=5x-2
1+2=5x
3/5=x right?

10^1=5x-2
10+2=5x
12/5=x
2 2/5=x right?

Notice the books answer is ]-0,75 (which is wrong, i agree), 1,6]

but i didnt get any answer 1,6. where is the mistake?
 
What is the problem? Initially it looked like the problem was where is

\(\displaystyle \sqrt(1-log(5x-2))\)

defined. Later it seemed to be changed to where is

1 / [e\(\displaystyle \sqrt(4x+3)\) \(\displaystyle \sqrt(1-log(5x-2)\)]

defined. Which, if either, of these should be the function under discussion?

Assuming it is the second one since it (almost) contains the first one we have three conditions to meet
(1) Because the argument of the log function must be positive
5x - 2 >0
or
x > 0.4
(2) because the argument of the square root must be non negative and because we can't divide by zero
1-log(5x-2) > 0
or, use log base 10 as the norm,
x < 2.4
(3) because the argument of the square root must be non negative
4 x + 3 >= 0
or
x >= -0.75

Putting all three of these together we have the domain of the second function above as (0.4, 2.4) because x>0.4 contains x>=-0.75.

Because (3) really adds nothing and in (2) we can have the function being zero if the proper function to consider is the first one, the domain of the first function is (0.4,2.4].
 
Last edited:
Good point! The answer in the book is wrong for the -0,75 part.

On the log part:
i take the log base = 10. Since nothing else is written there i assume its 10.
so
10^0=5x-2
1+2=5x
3/5=x right? ............. No - incorrect

10^1=5x-2
10+2=5x
12/5=x
2 2/5=x right?

Notice the books answer is ]-0,75 (which is wrong, i agree), 1,6]

but i didnt get any answer 1,6. where is the mistake?

for the lower limit:

5x - 2 > 0 → x > 0.4

I do not know why your book is providing the "wrong" answer.

Did you make sure you have copied the problem correctly and you are looking at the correctly labeled answer?

If answer to both the questions are "yes", then your book is simply wrong!

The correct answer of the "posted" problem is 0.4 < x ≤ 2.4
 
for the lower limit:

5x - 2 > 0 → x > 0.4

I do not know why your book is providing the "wrong" answer.

Did you make sure you have copied the problem correctly and you are looking at the correctly labeled answer?

If answer to both the questions are "yes", then your book is simply wrong!

The correct answer of the "posted" problem is 0.4 < x ≤ 2.4


It seems that the answer in the book just was wrong.... how long can a single problem cause trouble anyway....

Thankyou for the help and time!
 
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