A couple of questions... I don't understand... I'm breaking down because of no help!

Question

1) I don't get this-- if there is a video link with these kind of equations, please link me. If 3^5x+1 = 11 then x=? (Yes, the 3 is raised to 5x+1.)

2) This one too.. If f(x) = 1-x^3 and f^-1 is the inverse of f, how many solutions does the equation f(x) = f^-1(x) have? (Super confusing.....)


5) Which of the following numbers are in the range of f if f(x) = 2x-1 / x+1.
[part of Question 5] : I. -1
II. 1
III. 2

I kind of understand 5 but it's tough without a calculator (for me at least). How would you solve these without a calculator?

 

I'm stressing out so much! Someone please show me step by step-- I got these wrong on my test.

Answer 1

Any time you have X in the exponent, this is called an exponential equation.   In order to

solve for the X you have to get it down from the exponent.   In order to do this you change 

the equation from Exponential form to Logarithmic form:

 

Exponential Form:   y = X^b  

 

Logarithmic Form:    Log_x Y = b

 

The Y, X, and b have the same value in both of the Forms

 

Some times it is easier to work a problem when it is in one form

rather than in the other:

 

           3^{5x + 1} = 11          (exponential  form)

 

           5x + 1 = Log₃ 11  ( log form)

 

Once it is in log form, it is far easier to see how to work

Answer 2

I will help you with (1) and (5):

 

(1)

 

3^5x+1 = 11

 

I assume you have been exposed to logarithms (base 10 and ln?) in your class.  Take the logarithm on both sides of the equation and get

 

log(3^5x+1) = log(11)

 

Then using the properties of logarithms, get

 

(5x+1)log(3) = log(11)

 

 

5x+1 = log(11) / log(3)

 

x = [log(11) / log(3) - 1] / 5

 

(5)

 

f(x) = (2x-1) / (x+1)

 

Note x = -1 is not in the domain of f(x) because it gives zero in the denominator.

 

Test the three given candidates f(x) = -1, 1, and 2: 

 

f(x) = -1:

 

-1 = (2x-1) / (x+1) ⇒ -x-1 = 2x-1 ⇒ -3x = 0 ⇒ x = 0

Since x = 0 is in the domain, f(0) = -1 is in the range.

 

f(x) = 1:

 

1 = (2x-1) / (x+1) ⇒ x+1 = 2x-1 ⇒ -x = -2

x = 2

Since x = 2 is in the domain, f(2) = 1 is in the range.

 

f(x) = 2:

 

2 = (2x-1) / (x+1) ⇒ 2(x+1) = 2x-1 ⇒ 2x+2 = 2x-1 ⇒ 2 = -1 (FALSE!)

Since this is false, 2 cannot be in the range of f(x).

 

 

 

Answer 3

1)

 

You have an equation that has an unknown variable as an exponent.  Solve applying logarithms.  When solving these equations, you log both sides of the equation.  Then, you bring any exponents down as the coefficient of the log.  Then isolate the unknown variable using simple algebra.

 

 

2)

 

f^-1(x) is an inverse function.  Simply switch places with the x and y values from the original function.

 

x = 1 - y³

 

 

Then solve for y from this equation.

 

x - 1 = -y³

 

1 - x = y³

 

 

y = f^-1(x) = ³√(1 - x)   --->  inverse function

 

Now just equate the original function to the inverse function.

 

1 - x³ = ³√(1 - x)

 

Solve the bolded equation for x.

 

 

 

5)

 

To find the range, you need to find the horizontal asymptote.  The horizontal asymptote is   y=2.

 

So your range is in the interval (-∞, 2)∪(2, ∞). Also, note that f(-1) is undefined.