Solutions and full explanations to the set of college algebra multiple choice questions are presented.

9^{log9(4)} =
Solution
Exponential and log functions are inverse of each other. Hence
a^{loga(x)} = x , for all x real and positive.
and therefore
9^{log9(4)} = 4

3^{log3(5)} =
Solution
Since 5 is not in the domain of function log_{3}(x),
3^{log3(5)} is undefined

If f(x) = 2x^{2} + 8x  4, which of the follwoing is true?
A. The maximum value of f(x) is 4. 
B. The graph of f opens upward. 




C. The graph of f has no xintercept 
D. f is not a one to one function. 
Solution
f(x) is a quadratic function and its graph is a parabola that may be intercepted by horizontal lines at two points and therefore is not a one to one function. The answer is D

If f(x) = 5  2^{x}, then f^{ 1}(3) =
Solution
Find f^{ 1}(x) and then Find f^{ 1}( 3)
y = 5  2^{x} , given
x = 5  2^{y} , interchange x and y
2^{y} = 5  x , y = log_{2}(5  x) , solve for y
f^{ 1}(x) = log_{2}(5  x) , inverse function
f^{ 1}( 3) = log_{2}(5 + 3)
= log_{2}(2^{3}) = 3

If log_{x}(3) = 1/4, then x =
Solution
Rewrite the given equation in exponential form
log_{x}(3) = 1/4 if and only if x^{(1/4)} = 3
We now solve, for x, the exponential equation obtained above by raising both sides to the power 4.
(x^{(1/4)})^{ 4} = 3^{ 4}
x = 3^{ 4} = 81

If f(x) = x^{2} + 1, then f(x + 1) =
Solution
Substitute x by x + 1 in the formula of f(x) to obtain f(x + 1).
f(x + 1) =  (x + 1)^{ 2} + 1
Expand and simplify.
f(x + 1) =  x^{ 2}  2x  1 + 1 =  x^{ 2}  2x

If f(x) = x  4, then (f_{ o }f)(3) =
Solution
(f_{ o }f)(3) = f(f(3)) = f(3  4) = f(1) =  5

If ln(3x  2) = 1, then x =
Solution
Rewrite given equation in exponential form.
ln(3x  2) = 1 if and only if e^{ 1} = 3x  2
Solve e^{ 1} = 3x  2 for x.
x = (e + 2) / 3

The number of solutions of (x^{2} + 1)^{2} + 2(x^{2} + 1)  3 = 0 is equal to
Solution
Let u = x^{2} + 1 and rewrite the given equation in terms of u as follows
u^{ 2} + 2u  3 = 0
Factor and solve the above equation
(u + 3)(u  1) = 0
two solutions: u = x^{2} + 1 =  3 and u = x^{2} + 1 = 1
Equation x^{2} + 1 =  3 has no solutions. Solve the equation x^{2} + 1 = 1 for to get
x = 0.
The given equation has one solution.

If the graph of y = (x  2)^{2}  3 is translated 5 units up and 2 units to the right, then the equation of the graph obtained is given by
Solution
If the graph of y = f(x) is translated 5 units up, the equation of the new graph is given by
y = f(x) + 5
If the graph of y = f(x) + 5 is translated 2 units to the right, the equation of the new graph is given by
y = f(x  2) + 5 = ((x  2)  2)^{2}  3 + 5
= (x  4)^{2} + 2

If f(x) = e^{x}  2, then the range of f is given by the interval
A. (∞ , 2) 
B. (∞ , +∞) 




C. (2; , +∞) 
D. (∞ , +2) 

If f(x) = √(x  1) / (x^{2}  9), then the domain of f is given by the interval
A. (1 , +∞) 
B. (3 , +3) 




C. [1 , 3)U(3 , +∞) 
D. (3 , 3)U(3 , +∞) 

The number of points of intersections of the graphs of y = 2^{x} and y = x^{2} + 2 is equal to

If f(x) = ln(x + 1)  2, then f^{1}(x) =
A. e^{x + 1}  2

B. e^{x}  2





C. e^{x + 2}  2

D. e^{x + 2}  1


For all x real, √(x^{2} 4x + 4) =
A. x  2 
B. x + 2x + 2 




C. x  2 
D. x + 2 

The value of x that makes x^{2} + 6x + 13 maximum is equal to

e^{ln(3)  ln(2) + ln(1/x)} =
A. 3 / (2x) 
B. 3x/2 




C. 1 + 1/x 
D. 3/2  1/x 

If f(x) = (x  1) / (x + 2), then the range of f is given by the interval
A. (∞ , 2)U(2 , +∞) 
B. (∞ , 1)U(1 , +∞) 




C. (2; , +∞) 
D. (∞ , 1) 

ln((x  1)^{2}) = 2 ln(x  1) for all x in the interval
A. (∞ , +∞) 
B. [0 , +∞) 




C. (∞ , 1)U(1 , +∞) 
D. (1 , +∞) 

Let f(x) = x^{2} + 2x + 4. Which of the following statements is NOT true?
A. f(x) has a maximum value 
B. The graph of f is not a line 




C. The graph of f has no xintercepts. 
D. The graph of f has a yintercept. 

