Math  /  Algebra

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MHF4U Test \#3: Chapters 6-7 Name: \qquad Part A - Multiple Choice [K/U - 15 marks]
1. The function f(x)=(14)xf(x)=\left(\frac{1}{4}\right)^{x} passes through the point a. (1,4)(-1,4) b. (1,4)(1,-4) c. (1,4)(-1,-4)

Date: \qquad
2. The xx-intercept of the function y=6xy=6^{x} is a. 1 b. 6 c. 0 d. does not exist
3. The exponential function for the following data set is a. y=9xy=9^{x} b. y=(19)xy=\left(\frac{1}{9}\right)^{x} c. y=3xy=3^{x} d. y=(13)xy=\left(\frac{1}{3}\right)^{x} \begin{tabular}{|c|c|} \hline x\boldsymbol{x} & y\boldsymbol{y} \\ \hline 2 & 9 \\ \hline 3 & 27 \\ \hline 4 & 81 \\ \hline 5 & 243 \\ \hline \end{tabular}
4. Another way of writing 83=5128^{3}=512 is a. log8512=3\quad \log _{8} 512=3 b. log3512=8\log _{3} 512=8 c. log5128=3\log _{512} 8=3 d. log83=512\log _{8} 3=512
5. Another way of writing a=log2116a=\log _{2} \frac{1}{16} is a. a2=116\quad a^{2}=\frac{1}{16} b. (116)2=2\left(\frac{1}{16}\right)^{2}=2 c. a116=2a^{\frac{1}{16}}=2 d. 2a=1162^{a}=\frac{1}{16}
6. Evaluate log327\log _{3} 27. a. 0 b. 1 c. 2 d. 3
7. The equation of the vertical asymptote for the function y=3log(x+4)y=3 \log (x+4) is a. x=4x=-4 b. x=3x=3 c. x=4x=4 d. x=0x=0
8. The xx-intercept of the function y=4log(x+6)y=4 \log (x+6) is a. 4 b. -5 c. 5 d. 6
9. Express 10244\sqrt[4]{1024} as a power with a base of 2 . a. 2102^{10} b. 2252^{\frac{2}{5}} c. 2522^{\frac{5}{2}} d. 2342^{\frac{3}{4}}
10. Express 17417^{4} as a power with a base of 2 . a. log2174\log _{2} 17^{4} b. 2log2log172^{\frac{\log 2}{\log 17}} c. log217×log24\log _{2} 17 \times \log _{2} 4 d. 24log17log22^{\frac{4 \log 17}{\log 2}}

Studdy Solution

STEP 1

What is this asking? This question asks us to find the exponential function that fits the given data table. Watch out! Don't just look at the first data point; make sure the function works for *all* the points in the table!

STEP 2

1. Analyze the Data
2. Test Each Option

STEP 3

Let's look at how the y\boldsymbol{y} values change as x\boldsymbol{x} increases.
Notice that when x\boldsymbol{x} goes up by 1, the y\boldsymbol{y} value gets multiplied by a certain factor.
From x=2\boldsymbol{x=2} to x=3\boldsymbol{x=3}, y\boldsymbol{y} goes from **9** to **27**.
That's a factor of 27/9=327/9 = \boldsymbol{3}.
From x=3\boldsymbol{x=3} to x=4\boldsymbol{x=4}, y\boldsymbol{y} goes from **27** to **81**.
That's also a factor of 81/27=381/27 = \boldsymbol{3}.
This looks like an exponential function with a **base of 3**!

STEP 4

We can express this relationship as y=3somethingy = 3^{\text{something}}.
Since when x=2x = 2, y=9=32y = 9 = 3^2, and when x=3x = 3, y=27=33y = 27 = 3^3, it seems like the "something" is just xx!
So our function might be y=3xy = 3^x.

STEP 5

If x=2x = 2, y=92=81y = 9^2 = 81.
This doesn't match our table, where when x=2x = 2, y=9y = 9.
So, option (a) is **incorrect**.

STEP 6

If x=2x = 2, y=(1/9)2=1/81y = (1/9)^2 = 1/81.
Again, this doesn't match our table.
So, option (b) is **incorrect**.

STEP 7

Let's test a few points.
If x=2x = 2, y=32=9y = 3^2 = 9.
If x=3x = 3, y=33=27y = 3^3 = 27.
If x=4x = 4, y=34=81y = 3^4 = 81.
If x=5x = 5, y=35=243y = 3^5 = 243.
It matches all the points in the table!
So, option (c) is **correct**!

STEP 8

If x=2x = 2, y=(1/3)2=1/9y = (1/3)^2 = 1/9.
This doesn't match our table.
So, option (d) is **incorrect**.

STEP 9

The exponential function that represents the data is y=3xy = 3^x, which is option (c).

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