Problem Solving Drill – The Logarithmic Functions
Question No. 1 of 10
Instruction: (1) Read the problem statement and answer choices carefully
(2) Work the problems on paper as needed
(3) Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 1. If x>0,y>0, z>0 and t>0 which of the following is true:
Question #01
(A)
(B)
(C)
(D)
(E)
log2x + log2y + log2z
log2x + log2y + log2z
log2x + log2y + log2z
log2x + log2y + log2z
None of the above
-2
-2
-2
-2
log2t=
log2t=
log2t=
log2t=
log2(xy)
log2(xz)
log2(yz)
log2(x/t2)
A. Incorrect!
log2 x + log2 y + log2 z -2 log2 t= log2 (xy) Ù
log2 x + log2 y + log2 z -2 log2 t= log2 (x)+ log2 (y) Ù
log2 z -2 log2 t= 0 Ù
choose z=2 and t=2 and we get
log2z - 2log2 t= 1-2 = 1
So it is not always the case that log2 z - 2log2 t = 0
B. Incorrect!
log2 x + log2 y + log2 z -2 log2 t= log2 (xz) Ù
log2 x + log2 y + log2 z -2 log2 t= log2 (x)+ log2 (z) Ù
log2y -2 log2 t= 0 Ù
choose y=2 and t=2 and we get
log2y - 2log2 t= 1-2 = -1
So it is not always the case that log2 y - 2log2 t = 0
Feedback on
Each Answer
Choice
C. Incorrect!
log2 x + log2 y + log2 z -2 log2 t= log2 (yz) Ù
log2 x + log2 y + log2 z -2 log2 t= log2 (y)+ log2 (z) Ù
log2 x-2 log2 t= 0 Ù
choose x=2 and t=2 and we get
log2x - 2log2 t= 1-2 = -1
So it is not always the case that log2 x - 2log2 t = 0
D. Incorrect!
log2 x + log2 y + log2 z -2 log2 t= log2(x/t2) Ù
log2 x + log2 y + log2 z -2 log2 t= log2 (x)-2log2 (t) Ù
log2y+ log2 z= 0 Ù
choose y=2 and z=2 and we get
log2y+ log2 z= 1+1 = 2
So it is not always the case that log2 y +log2 z = 0
E. Correct!
Because A,B,C,D, are all incorrect.
Solution
log2 x + log2 y + log2z -2 log2 t=
log2 (xy) + log2 z -2 log2 t2 =
log2 (xyz) - 2 log2 t =
log2 (xyz) - log2 (t2) =
log2 (xyz / t2 )
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Question No. 2 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 2. Which of the following is true:
Question #02
A.
B.
C.
D.
E.
log2 1 =
log2 1 =
log2 1 =
log2 1 =
None of
2
3
4
5
the above
A. Incorrect!
2= log2 22 = log2 4 and log is an increasing function if the base is greater than 1, so log2 4
› log2 1
B. Incorrect!
3= log2 23 = log2 8 and log is an increasing function if the base is greater than 1, so log2 8
Feedback on
Each Answer
Choice
› log2 1
C. Incorrect!
4= log2 24 = log2 16 and log is an increasing function if the base is greater than 1, so log2 16
› log2 1
D. Incorrect!
5= log2 25 = log2 32 and log is an increasing function if the base is greater than 1, so log2 32
E. Correct!
Because A,B,C,D, are all incorrect. log2 1 = log2 20 = 0 . log2 2= 0
log2 1 = log2 20 = 0 . log2 2= 0
Solution
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› log2 1
Question No. 3 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 3.
If A > 0, which of the following is true:
Question #03
(A)
(B)
(C)
(D)
(E)
log2 x +
log2 x +
log2 x +
log2 x +
None of
n = log2 x, for all n Є R
n = log3 x + n, for all n Є R
n = log2 y + n , for all y > 0
n = log2 (xn)
the above
A. Incorrect!
log2 x + n = log2 x
n=0
B. Incorrect!
log2 x + n = log3 x + n
log2 x = log3 x
2=3 and this is incorrect.
C. Incorrect!
Feedback on
Each Answer
Choice
log2 x + n = log2 y + n
log2 x = log2 y Ù x = y and this is not always correct.
D. Incorrect!
log2 x + n = log2 (xn)
log2 x + n = log2 (x) + log2(n)
n = log2(n) and this is incorrect if we pick n = 2 for example.
E. Correct!
None of the above is correct, so E is correct.
Refer to individual response for detailed solution above
Solution
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Question No. 4 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 4:
If x > 0, which of the following is true?
Question #04
(A)
(B)
(C)
(D)
(E)
ln (5x+1) = ln (5x+3) for at least one value of x > 0
ln (5x+1) = ln (4x+3) for at least two values of x > 0
ln (5x+1) is always a positive number if x > -1/5
ln (5x+1) does not exist if x = 3
None of the above
A. Incorrect!
ln (5x+1) = ln (5x+3) Ù 5x+1 = 5x+3 Ù 1 = 3
B. Incorrect!
ln (5x+1) = ln (4x+3) Ù 5x+1 = 4x+3 Ù x = 2 so there are no 2 values of x
for which the equality holds
Feedback on
Each Answer
Choice
C. Incorrect!
Let x=0, ln(5x+1)=ln(1)=0 ; for -1/5 < x< 0, ln(5X+1) < 0
D. Incorrect!
If x=3, ln(5x+1)=ln(16)=2.77
E. Correct!
Because A,B,C,D, are all incorrect.
Refer to each answer response for solutions.
As none of the answers A to D are correct, E must be correct.
Solution
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Question No. 5 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 5. Which of the following is true?
Question #05
(A)
(B)
(C)
(D)
(E)
log2 4 +
log2 4 +
log2 4 +
log2 4 +
None of
log2 8 = 2
log2 8 = 3
log2 8 = 4
log2 8 = 6
the above
A. Incorrect!
log2 4 + log2 8
=log2 22 + log2 23
=2log2 2 + 3log2 2
=2+3=5
log2 4 + log2 8 = 5
B. Incorrect!
log2 4 + log2 8
=log2 22 + log2 23
=2log2 2 + 3log2 2
=2+3=5
Feedback on
Each Answer
Choice
log2 4 + log2 8 = 5
C. Incorrect!
log2 4 + log2 8
=log2 22 + log2 23
=2log2 2 + 3log2 2
=2+3=5
log2 4 + log2 8 = 5
D. Incorrect!
log2 4 + log2 8
=log2 22 + log2 23
=2log2 2 + 3log2 2
=2+3=5
log2 4 + log2 8 = 5
E. Correct!
None of the above is correct so the statement of E is correct.
log2 4 + log2 8
log2 22 + log2 23
Solution
2log2 2 + 3log2 2
2+3=5
log2 4 + log2 8 = 5
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Question No. 6 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 6. If x > 0, which of the following is true?
Question #06
(A)
(B)
(C)
(D)
(E)
log2 x +
log2 x +
log2 x +
log2 x +
None of
log2 (x+3)
log2 (x+3)
log2 (x+3)
log2 (x+3)
the above
=
=
=
=
2
2
2
2
has
has
has
has
0
3
4
5
solutions
solutions
solutions
solutions
A. Incorrect!
We can see that x = 1 is a solution for the given equation as
log2 1 + log2 (1+3) = 0 + log2 4 = 2
B. Incorrect!
We can manipulate the left hand side of the equation as follows:
log2 x + log2 (x+3) = log2(x*(x+3)), so x can only be a solution
of the given equation if x*(x+3) = 22 = 4 and a quadratic equation
has at most 2 solutions.
C. Incorrect!
Feedback on
Each Answer
Choice
We can manipulate the left hand side of the equation as follows:
log2 x + log2 (x+3) = log2(x*(x+3)), so x can only be a solution
of the given equation if x*(x+3) = 22 = 4 and a quadratic equation
has at most 2 solutions.
D. Incorrect!
We can manipulate the left hand side of the equation as follows:
log2 x + log2 (x+3) = log2(x*(x+3)), so x can only be a solution
of the given equation if x*(x+3) = 22 = 4 and a quadratic equation
has at most 2 solutions.
E. Correct!
None of the above is correct so the statement of E is correct.
log2 x + log2 (x+2) = 3 Ù
log2 (x*(x+2)) = 3
Ù
x *(x+2) = 8
x2+2x-8=0
x1, 2 =
Solution
− 2 ± 4 + 4 ∗8 − 2 ± 6
=
2
2
The quadratic equation has 2 real solutions, still only the positive one is good
as the logarithm log2x is only defined for x>0.
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Question No. 7 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 7. Which of the following is true?
(A)
(B)
(C)
(D)
(E)
Question #07
log (x+1) +
log (x+1) +
log (x+1) +
log (x+1) +
None of the
log (2-x)
log (2-x)
log (2-x)
log (2-x)
above
is
is
is
is
well
well
well
well
defined
defined
defined
defined
if
if
if
if
and
and
and
and
only
only
only
only
if
if
if
if
x
x
x
x
>0
>2
=4
Є (0,2)
A. Incorrect!
Choose x=3, 2-x =-1 and log (2-x) is invalid.
B. Incorrect!
Choose x=3, 2-x =-1 and log (2-x) is invalid.
Feedback on
Each Answer
Choice
C. Incorrect!
Choose x=4, 2-x =-2 and log (2-x) is invalid.
D. Incorrect!
Choose x=-1/2, x+1=1/2 and 2-x=5/2
For x=-1/2, log (x+1) + log (2-x)=log(1/2)+log(5/2)=log(5/4)
E. Correct!
None of the above is correct so the statement of E is correct.
The expression is well defined if both log (x+1) and log (2-x) are well defined, so we need
X+1> 0 Ù x>-1
and
2-x >0 Ù x<2
or
x ∈ (−1,2)
Solution
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Question No. 8 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 8. Which of the following functions is the reflection of the function
f: R+ÆR f(x)=log2(x) ?
(A)
(B)
(C)
(D)
(E)
Question #08
g: R-ÆR g(x)=x
g: R-ÆR g(x)=1
g: R-ÆR g(x)=0
g: R-ÆR g(x)=-x
None of the above
A. Incorrect!
For
g : R− → R
to be the reflection of the function f around the y axis, the following must be true:
g(x) = f(-x)
For x = 2, g(2) = 2 and log22=1, so g is not the reflection of f around the y axis.
B. Incorrect!
For
g : R− → R
to be the reflection of the function f around the y axis, the following must be true:
g(x) = f(-x)
For x = 4, g(2) = 1 and log24=2, so g is not the reflection of f around the y axis.
Feedback on
Each Answer
Choice
C. Incorrect!
For
g : R− → R
to be the reflection of the function f around the y axis, the following must be true:
g(x) = f(-x)
For x = 4, g(2) = 0 and log24=2, so g is not the reflection of f around the y axis.
D. Incorrect!
For
g : R− → R
to be the reflection of the function f around the y axis, the following must be true:
g(x) = f(-x)
For x = -2, g(-2) = 2 and log2(2)=1, so g is not the reflection of f around the y axis.
E. Correct!
None of the above is correct so the statement of E is correct.
For
g : R− → R
to be the reflection of the function f around the y axis, the following must be true:
g(x) = f(-x) or
g(x) = log2(-x)= log2(1/x)
Solution
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Question No. 9 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 9. Which of the following is true:
(A) x=2 is a solution for log3(x+1)=0
(B) x=8 is a solution for log3(x+1)=3
(C) x=2 is a solution for log3(x+1)=1
(D) x=-1 is a solution for log3(x+1)=0
(E) None of the above
Question #09
A. Incorrect!
If x=2 , x+1=3
log3(x+1) = log33 = 1 ≠ 0
B. Incorrect!
If x=8 , x+1=9 = 32
log3(x+1) = log332= 2 ≠ 3
Feedback on
Each Answer
Choice
C. Correct!
If x=2 , x+1=3
log3(x+1) = log33 = log3(31)= 1
D. Incorrect!
If x=-1, log3(0) = undefined
E. Incorrect!
There is one correct answer above.
If x=2
Then x+1=3
Log3(x+1) = log33 = log3(31)= 1
Solution
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Question No. 10 of 10
Instruction: (1) Read the problem statement and answer choices carefully (2) Work the problems on paper as needed (3)
Pick the answer (4) Go back to review the core concept tutorial as needed.
Question 10. Which of the solution of :x2ln(x)+1+e2=(1+e2)x(ln(x)-1) for x>0
Question #10
(A)
(B)
(C)
(D)
(E)
x=1
x=e
x=3
x=2e
None of the above
A. Correct!
The equation can be simplified to
x 2 + xe 2 − (1 + e 2 ) = 0 . So the solution of this algebra equation is
x=1. Thus. the statement of A is correct.
B. Incorrect!
The equation can be simplified to
x 2 + xe 2 − (1 + e 2 ) = 0 . X=e is not the solution of this equation.
Thus. the statement of B is incorrect.
C. Incorrect!
Feedback on
Each Answer
Choice
The equation can be simplified to
x 2 + xe 2 − (1 + e 2 ) = 0 . X=3 is not the solution of this equation.
Thus. the statement of C is incorrect.
D. Incorrect!
The equation can be simplified to
x 2 + xe 2 − (1 + e 2 ) = 0 . X=2e is not the solution of this equation.
Thus. the statement of D is incorrect.
E. Incorrect!
Because A is correct. So this statement is wrong.
Let’s simplify the various terms of the equation:
x 2 ln( x ) +1 = x 2 ln( x ) x = ( x ln( x ) ) 2 x = (1) 2 x = x
x (ln( x ) −1) =
x ln( x ) 1
=
x
x
So, the equation becomes:
Solution
x + e 2 = (1 + e 2 )
1
x
Ù
x 2 + xe 2 − (1 + e 2 ) = 0
x1, 2 =
Ù
− e 2 ± e 4 + 4(1 + e 2 )
2
Because x>0, the solution is
(
− e 2 + e 4 + 4(1 + e 2 ) − e 2 + e 2 + 2
x1 =
=
2
2
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)
2
=
− e2 + e2 + 2
=1
2
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