Calculus problem?

Question

1) Solve the equation:
In(Inx)=1

2) What's the domain and range of f(x)=square root [4-3(x^2)]

Answer 1

1) ln(ln(x)) = 1
e^1 = ln(x) (convert to exponential form)
e = ln(x)
x = e^e = 15.154

2) The domain is the set of x values that makes 4 - 3x^2 >= 0
So solve 4 - 3x^2 = 0
4 = 3x^2
x^2 = 4/3
x = +or- 2sqrt(3)/3
The x values that satisfy the inequality are in the interior of that interval, so the domain is [-2sqrt(3)/3, 2sqrt(3)/3]

The range is [0,2] (graphed it.)

Answer 2

1:

In(Inx)=1

e^ (In(Inx)) = e ^ 1

lnx = e

e^ lnx = e ^e

x = e^ e

2)What's the domain and range of f(x)=square root [4-3(x^2)]

assuming "reals", we need
4 - 3(x^2) >= 0
3(x^2) <= 4
x^2 <= 4/3
-sqrt(4/3) <= x <= sqrt(4/3)

just guessing for the 'domain' ...

range ?? nahh .. you do the math

Answer 3

1) ln(lnx)=1
(take the exponential of both the sides to eliminate the first natural log function), we get
lnx=e
(again take the exponential of both sides), we get
x=e^e
The answer is e raised to power of e
2) For f(x) to exist, 4-3(x^2) >=0;
thus, 4>=3x^2
x^2<=(4/3)
thus, Domain: x is between -sqrt(4/3) and +sqrt(4/3)
Range: -2 to +2

Answer 4

1)
ln(lnx) = 1
e^(ln(lnx)) = e^1 raise both sides
ln x = e simplify
e^(lnx) = e^e raise both sides again
x=e^e simplify

2).
sqr(4-3x^2)
Domain: 4-3x^2 must be positive
Set 4-3x^2=0
-3x^2=-4
3x^2=4
x^2=4/3
x= plus or minus sqr(4/3)

Domain: x extends from -sqr(4/3) to +sqr(4/3)

Range:
There is a max at 0 of 2 and a min at sqr(4/3) at about .070
Therefore y extends from .070 to 2


Domain

Answer 5

1)
ln(lnX) = 1
e ^ ln(lnx) = e^1
ln x = e
x = e ^ e
x = 15.154

2)

F(X) = √(4 -3x²)
4-3x² >=0
4 >=3x²
X² <= 4/3
-2/√3 <= x <= 2/√3
Domain of F(X) = [-2 / √ 3 , 2 / √ 3 ]
Range = [-2 , 2]

Answer 6

1.x=e^e where e=2.303 approximately
2.4-3x^2>=0 domain=[-4/(3^0.5),4/(3^0.5)]
range any positive real number

Answer 7

Do you mean ln(x)=1? in that case to get rid of the ln you need to raise both sides to the power of e. Thus x=e...