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Make up DOL Mixed Problems

Total questions: 10

Worksheet time: 22mins

Name
Class
Date
1.

Find the mistake

step 1 f(x)=−3log⁡(−x−6)+12step\ 1\ f\left(x\right)=-3\log\left(-x-6\right)+12  

step 2    y=−3log⁡(−x−6)+12step\ 2\ \ \ \ y=-3\log\left(-x-6\right)+12

step 3     x=−3log⁡(−y−6)+12step\ 3\ \ \ \ \ x=-3\log\left(-y-6\right)+12

step 4     x−12=−3log⁡(−y−6)step\ 4\ \ \ \ \ x-12=-3\log\left(-y-6\right)

step 5    −13x−4=log⁡(−y−6)step\ 5\ \ \ \ -\frac{1}{3}x-4=\log\left(-y-6\right)

step 6    10−13x−4=−y−6step\ 6\ \ \ \ 10^{-\frac{1}{3}x-4}=-y-6

step 7    f−1(x)=−10−13x−4−6step\ 7\ \ \ \ f^{-1}\left(x\right)=-10^{-\frac{1}{3}x-4}-6

a)

Step 2 

b)

step 3 

c)

step 4 

d)

step 5

e)

To see all the steps scroll up and down.

2.

Find the inverse of f(x)=−4(5)x−7+12f\left(x\right)=-4\left(5\right)^{x-7}+12  

a)

f−1(x)=log⁡5(−14x+3)+7f^{-1}\left(x\right)=\log_5\left(-\frac{1}{4}x+3\right)+7  

b)

f−1(x)=7log⁡5(−14x+3)f^{-1}\left(x\right)=7\log_5\left(-\frac{1}{4}x+3\right)  

c)

f−1(x)=log⁡5(x−3)−7f^{-1}\left(x\right)=\log_5\left(x-3\right)-7  

3.
In the general log function 
y = a*logb(x-h)+k, what letter tells us if the function has moved left or right?
a)
a
b)
b
c)
h
d)
k
4.

Select all the transformations that apply.

y = 5log2 (x)

a)

Vertical stretch

b)

Vertical Compression

c)

Horizontal compression

d)

Horizontal stretch

5.
What is the transformation?
a)
Vertical Translation up 2 and Horizontal translation right 1
b)
Vertical Translation down 2 and Horizontal Translation left 1
c)
Horizontal Translation left 2 and Vertical Translation up 1
d)
Horizontal Translation right 2 and Vertical Translation down 1
6.
What is the transformation?
a)
Right 3 and Up 7
b)
Left 3 and Up 7
c)
Right 3 and Down 7
d)
Left 3 and Down 7
7.

Log32x=5Log_32x=5  can be rewritten as...

a)

2x=352x=3^5  

b)

5=32x5=3^{2x}  

c)

2x=532x=5^3  

8.

Log4x=2Log4x=2  can be rewritten as...

a)

4x=1024x=10^2  

b)

2=104x2=10^{4x}  

c)

10=46x10=4^{6x}  

9.

Which of the following is the correct first two steps when solving for x

log⁡7(x−3)+log⁡7(x+1)=1\log_7\left(x-3\right)+\log_7\left(x+1\right)=1

a)

log⁡7(x2−2x−3)=1\log_7\left(x^2-2x-3\right)=1 x2−2x−3=71x^2-2x-3=7^1

b)

x2−2x−3=1x^2-2x-3=1 x2−2x−2=0x^2-2x-2=0

c)

log⁡7(x2−2x−3)=1\log_7\left(x^2-2x-3\right)=1 7x2−2x−3=17^{x^2-2x-3}=1

d)

log⁡7(1)=x2−2x−3\log_7\left(1\right)=x^2-2x-3 log⁡7(x2−2x−3)=1\log_7\left(x^2-2x-3\right)=1

10.

Solve for x. log⁡8(x)+log⁡8(x+2)=1\log_8\left(x\right)+\log_8\left(x+2\right)=1  

a)

2, 13

b)

13

c)

2

d)

Hint: When you add logs into one log, multiply the inside. Then rewrite the log as a power to solve.