Understanding exponents - GMAT Quantitative

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Question

The first and third terms of a geometric sequence are and $K ^{4}$ , in that order. Give the eighth term.

(Assume is positive.)

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Answer

Let be the common ratio of the geometric sequence. Then the third term is $r^{2}$ times the first, so

$r^{2} = \frac{K^{4}}{K} = K^{3}$

and

$r = \sqrt{K^{3}} = K^ { \frac{3}{2}}$ .

The eighth term of the sequence is

$a_{8} = a_{1} \cdot r ^{7}$

$= K \cdot (K^{\frac{3}{2}} )^{7}$

$= K \cdot K^{\frac{21}{2}}$

$= K^{\frac{23}{2}}$

$= K^{11 \frac{1}{2}}$

$= K^{11} \cdot K^{ \frac{1}{2}}$

$= K^{11 } \sqrt{K}$

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Question

The first two terms of a geometric sequence are and $C ^{3}$ , in that order. Give the tenth term.

(Assume is positive.)

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Answer

The common ratio of the geometric sequence can be found by dividing the second term by the first:

$\frac{C^{3}}{C} = C^{2}$

The tenth term of the sequence is therefore

$a_{10} = a_{1}r^{10-1}=C \cdot \left ( C^{2}\right ) ^{9} = C \cdot C^{18} = C^{19}$

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Question

$7^{3x+2} = 49 ^{y - 12 }$

Express in terms of .

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Answer

$7^{3x+2} = 49 ^{y - 12 }$

$7^{3x+2} = \left (7^{2} \right ) ^{y - 12 }$

$7^{3x+2} = 7^{2 \cdot (y - 12 )}$

$7^{3x+2} = 7^{2 y - 24}$

$\frac{1}{2}\cdot 2y = \frac{1}{2}\cdot (3x+2 6)$

$y = \frac{3}{2} x+13$

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Question

$4^{t} = \frac{2^{x+3}}{8^{y - 5}}$

Which of the following is a true statement?

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Answer

$4^{t} = \frac{2^{x+3}}{8^{y - 5}}$

$(2^{2}) ^{t} = \frac{2^{x+3}}{(2^{3}) ^{y - 5}}$

$2^{2t} = \frac{2^{x+3}}{2^{3(y-5)}}$

$2^{2t} = \frac{2^{x+3}}{2^{3y-15}}$

$2^{2t} = 2^{(x+3)- (3y-15)}$

$\frac{1}{2} \cdot 2t= \frac{1}{2} \cdot ( x - 3y+18 )$

$t= \frac{1}{2} x - \frac{3}{2} y+9$

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Question

$0.25 ^{t} = 2 ^{3y}$

Express in terms of .

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Answer

$0.25 ^{t} = 2 ^{3y}$

$\left ( \frac{1}{4} \right )^{t} = 2 ^{3y}$

$\left ( 2 ^{-2 }\right )^{t} = 2 ^{3y}$

$2 ^{-2t} = 2 ^{3y}$

$\frac{-2t }{-2 }= \frac{3y}{-2}$

$t = -\frac{3}{2}y$

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Question

Simplify: $\frac{x^6y^{-4}z^0}{x^{-5}y^{10}z^{-6}}$

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Answer

To solve, we must first simplify the negative exponents by shifting them to the other side of the fraction:

$\frac{x^6y^{-4}z^0}{x^{-5}y^{10}z^{-6}} = \frac{x^6x^5z^0z^6}{y^{10}y^4}$

Then we can simply the multiplied like bases by adding their exponents:

$\frac{x^6x^5z^0z^6}{y^{10}y^4} = \frac{x^{11}z^6}{y^{14}}$

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Question

Which of the following is equal to $2 + \log 6$ ?

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Answer

$2 + \log 6 = \log \left ( 10^{2} \right ) + \log 6 = \log 100 + \log 6 = \log\left ( 100 \cdot 6 \right ) = \log 600$

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Question

Divide:

$\left (56x^{3} +21x ^{2} -63x + 77 \right )\div 7 x ^{2}$

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Answer

$\left (56x^{3} +21x ^{2} -63x + 77 \right )\div 7 x ^{2}$

$= \frac{ 56x^{3} }{7 x ^{2}} + \frac{ 21x ^{2} }{7 x ^{2}}- \frac{ 63x }{7 x ^{2}}+ \frac{ 77 }{7 x ^{2}}$

$= \frac{ 56 }{7 } x^{3-2} + \frac{ 21 }{7 }- \frac{ 63 }{7}\cdot \frac{1}{ x ^{2-1}}+ \frac{ 77 }{7} \cdot \frac{1}{ x ^{2}}$

$= 8 x +3 -\frac{9}{ x }+ \frac{ 11 }{x ^{2}}$

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Question

. Order from least to greatest: $x^{-1}, x^{-2}, x^{-3}, x^{-4}$ .

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Answer

If is negative, then:

Even powers $x^{2}$ and $x^{4}$ are positive, with $x^{2} = |x| ^{2}$ and $x^{4} = |x| ^{4}$ .

Since $x \in \left (-1, 0 \right )$ , ,

It follows that $|x|^{4}<| x|^{2}$ , and $x^{4}<x^{2}$ .

If $x^{M }< x^{N}$ , then $\frac{1}{x^{M }}> \frac{1}{x^{N}}$ - that is, $x^{-M}> x^{-N}$ . So changing the signs of the exponents reverses the order. As a result,

$x^{-2} < x^{-4}$ .

Odd powers and $x^{3}$ are negative, with $x =- |x| ^{1}$ and $x^{3} = - |x| ^{3}$ .

$|x| ^{3} < |x|^{1}$ , so $-|x| ^{3} > -|x|^{1}$ , and $x^{1}< x^{3}$ .

As before, changing their exponents to their opposites reverses the order:

$x^{-3}< x^{-1}$

Setting the negative numbers less than the positive numbers:

$x^{-3}< x^{-1}<x^{-2}< x^{-4}$ .

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Question

$\frac{6^{3}}{36} + \frac{3^{68}}{3^{67}}=$

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Answer

$\frac{6^{3}}{36} = \frac{6^{3}}{6^{2}} = 6$

$\frac{3^{68}}{3^{67}} = 3^{68-67} = 3$

Putting these together,

$\frac{6^{3}}{36} + \frac{3^{68}}{3^{67}}= 6 + 3 = 9$

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Question

$\dpi{100} \small 3x^{4}\times x^{2}+x^{2}-x =$

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Answer

$\dpi{100} \small 3x^{4}\times x^{2} =3x^{6}$

Then, $\dpi{100} \small 3x^{4}\times x^{2}+x^{2}-x = 3x^{6}+x^{2}-x = x(3x^{5}+x-1)$

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Question

$4^{\frac{3}{2}} + 27^{\frac{2}{3}} =$

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Answer

$4^{\frac{3}{2}}=(4^{\frac{1}{2}})^{3} = 2^{3} = 8$

$27^{\frac{2}{3}}=(27^{\frac{1}{3}})^{2} = 3^{2} = 9$

Then putting them together, $4^{\frac{3}{2}} + 27^{\frac{2}{3}} = 8 + 9 = 17$

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Question

If , what is in terms of ?

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Answer

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

So , and .

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Question

Which of the following expressions is equivalent to this expression?

$\small \frac{x^{-\frac{1}{2}}}{x^{\frac{1}{3}}}$

You may assume that $\small x > 0$ .

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Answer

$\small \small \small \small \small \small \small \small \small \frac{x^{-\frac{1}{2}}}{x^{\frac{1}{3}}} = x^{-\frac{1}{2} -{\frac{1}{3}} } = x^{-\frac{2}{6} -{\frac{3}{6}} }= x^{-\frac{5}{6} }= \frac{1}{ x^{\frac{5}{6}}}$

$\small \small \small = \frac{1}{\sqrt[6]{x^{5}}}= \frac{\sqrt[6]{x}}{\sqrt[6]{x^{5}}\cdot \sqrt[6]{x}}= \frac{\sqrt[6]{x}}{x}$

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Question

Simplify the following expression without a calculator:

$(\frac{9x^\frac{-5}{2}y^\frac{7}{4}}{x^\frac{3}{2}y^\frac{-1}{4}})^\frac{3}{2}$

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Answer

The 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, $a^\frac{-5}{2} = \frac{1}{a^\frac{5}{2}}$ So using this, we simplify: $(\frac{9x^\frac{-5}{2}y^\frac{7}{4}}{x^\frac{3}{2}y^\frac{-1}{4}})^\frac{3}{2} = (\frac{9(y^\frac{7}{4})(y^\frac{1}{4})}{(x^\frac{5}{2})(x^\frac{3}{2})})^\frac{3}{2}$

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 $(\frac{9y^2}{x^4})^\frac{3}{2}$ .

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, $(a^2)^4 = a^{(24)}= a^8$ . The other rule we must know that is an exponent of one half is the same as taking the square root. So for the $9^{3/2}=(\sqrt9)^3$ So using these rules, $(\frac{9y^2}{x^4})^\frac{3}{2} = \frac{9^{3/2}(y^2)^{3/2}}{(x^4)^{3/2}} = \frac{(\sqrt9)^3y^{(23/2)}}{x^{(4*3/2)}} = \frac{3^3y^3}{x^6}= \frac{27y^3}{x^6}$

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Question

Which of the following numbers is in scientific notation?

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Answer

The number in front must have an absolute value greater than or equal to 1, and less than 10. Of these choices, only $6.7 \times 10^{-9}$ qualifies.

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Question

If , what does equal?

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Answer

We can use the fact that $(x^a)^b=x^{a*b}$ to see that

Since , we have

.

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Question

Solve for :

$\left (3 ^{4} \right )^{N}\cdot 3^{5} =\frac{1}{27}$

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Answer

$\left (3 ^{4} \right )^{N}\cdot 3^{5} =\frac{1}{27}$

$3 ^{4\cdot N} \cdot 3^{5} =3 ^{-3 }$

$3 ^{4\cdot N+5} =3 ^{-3 }$

The left and right sides of the equation have the same base, so we can equate the exponents and solve:

$4N\div 4 = -8 \div 4$

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Question

Rewrite as a single logarithmic expression:

$3 + \log_{3} x - 2\log_{3} y$

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Answer

First, write each expression as a base 3 logarithm:

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

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

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

$3 + \log_{3} x - 2\log_{3} y$

$=\log_{3}27 + \log_{3} x - \log_{3} y^{2}$

$=\log_{3}\left (27 \cdot x \right )- \log_{3} y^{2}$

$=\log_{3}\left (27 \cdot x \right \div y^{2} ) = \log_{3}\left (\frac{27 x }{y^{2}}\right )$

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Question

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

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Answer

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.

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