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QUESTION: 1

Solution:

Putting these together,

QUESTION: 2

Solution:

Then putting them together,

QUESTION: 3

Which of the following expressions is equivalent to this expression?

You may assume that x>0.

Solution:

QUESTION: 4

Simplify the following expression without a calculator:

Solution:

he easiest way to simplify is to work from the inside out. We should first get rid of the negatives in the exponents. Remember that variables with negative exponents are equal to the inverse of the expression with the opposite sign. For example

So using this, we simplify:

Now when we multiply variables with exponents, to combine them, we add the exponents together. For example, (a^{2})(a^{2})=a^{2+2}=a^{4} Doing this to our expression we get it simplified to

The next step is taking the inside expression and exponentiating it. When taking an exponent of a variable with an exponent, we actually multiply the exponents. For example, (a^{2})^{4}=a^{(2∗4)}=a^{8}.

The other rule we must know that is an exponent of one half is the same as taking the square root. So for the So using these rules,

QUESTION: 5

Rewrite as a single logarithmic expression:

3+log_{3}x−2log_{3}y

Solution:

First, write each expression as a base 3 logarithm:

3=log_{3}27 since 3^{3}=27

2log_{3}y=log_{3}y^{2}

Rewrite the expression accordingly, and apply the logarithm sum and difference rules:

3+log_{3}x−2log_{3}y

=log_{3}27+log_{3}x−log_{3}y^{2}

=log_{3}(27⋅x)−log_{3}y^{2}

QUESTION: 6

If 3^{p}+3^{p}+3^{p}=3^{q}, what is p in terms of q?

Solution:

We have 3^{p}+3^{p}+3^{p}=3∗3^{p}=3^{p+1}=3q.

So p+1=q, and p=q−1.

QUESTION: 7

What are the last two digits, in order, of 6^{789} ?

**Possible Answers:**

Solution:

Inspect the first few powers of 6; a pattern emerges.

6^{1}=6

6^{2}=36

6^{3}=216

6^{4}=1,296

6^{5}=7,776

6^{6}=46,656

6^{7}=279,936

6^{8}=1,679,616

6^{9}=10,077,696

6^{10}=60,466,176

As you can see, the last two digits repeat in a cycle of 5.

789 divided by 5 yields a remainder of 4; the pattern that becomes apparent in the above list is that if the exponent divided by 5 yields a remainder of 4, then the power ends in the diigts 96.

QUESTION: 8

Which of the following expressions is equal to the expression

Solution:

Use the properties of exponents as follows:

(x^{3})^{4}⋅(x^{3})^{5}

=x^{3⋅4 }⋅ x^{3⋅5}

=x^{12}⋅ x^{15}

=x^{12+15}

=x^{27}

QUESTION: 9

Simplify:

Solution:

Apply the power of a power principle twice by multiplying exponents:

QUESTION: 10

Divide:

(56x^{3}+21x^{2}−63x+77)÷7x^{2}

Solution:

QUESTION: 11

Solve for N:

Solution:

(1/25)^{N }⋅ 5^{6}=125

(5^{−2})^{N }⋅ 5^{6}=53

5^{−2⋅N }⋅ 5^{6}=53

5^{−2N }+ 6=53

−2N+6=3

−2N+6−6=3−6

−2N=−3

−2N÷(−2)=−3÷(−2)

N=3/2

QUESTION: 12

Which of the following is equal to log8+log5−log4 ?

Solution:

log8+log5−log4

=log(8⋅5)−log4

=log40−log4

=log(40÷4)

=log10=1

QUESTION: 13

Which of the following is equal to 4log3−2log6 ?

Solution:

4log3−2log6

=log(3^{4})−log(6^{2})

=log81−log36

=log81/36

=log9/4

=log 9− log 4

QUESTION: 14

Solution:

QUESTION: 15

Which of the following is true about log_{5}1,000 ?

Solution:

5^{4}=625;5^{5}=3,125

Therefore,

log_{5 }625=4;log^{5 }3,125=5

log_{5}625<log_{5}1,000<log_{5}3,125

4<log_{5}1,000<5

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