How to find patterns in exponents - SAT Math

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Question

If is the complex number such that , evaluate the following expression:

$i^{25}-i^{23}+i^{21}-i^{19}+i^{17}-i^{15}.\ .\ .+i$

Answer

The powers of i form a sequence that repeats every four terms.

i1 = i

i2 = -1

i3 = -i

i4 = 1

i5 = i

Thus:

i25 = i

i23 = -i

i21 = i

i19= -i

Now we can evalulate the expression.

i25 - i23 + i21 - i19 + i17..... + i

= i + (-1)(-i) + i + (-1)(i) ..... + i

= i + i + i + i + ..... + i

Each term reduces to +i. Since there are 13 terms in the expression, the final result is 13i.

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Question

If ax·a4 = a12 and (by)3 = b15, what is the value of x - y?

Answer

Multiplying like bases means add the exponents, so x+4 = 12, or x = 8.

Raising a power to a power means multiply the exponents, so 3y = 15, or y = 5.

x - y = 8 - 5 = 3.

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Question

What digit appears in the units place when $2^{102}$ is multiplied out?

Answer

This problem is quite simple if you recall that the units place of powers of 2 follows a simple 4-step sequence.

Observe the first few powers of 2:

21 = 2, 22 = 4, 23 = 8, 24 = 16, 25 = 32, 26 = 64, 27 = 128, 28 = 256 . . .

The units place follows a sequence of 2, 4, 8, 6, 2, 4, 8, 6, etc. Thus, divide 102 by 4. This gives a remainder of 2.

The second number in the sequence is 4, so the answer is 4.

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Question

If $x^7y^8z^{10} < 0$ , then which of the following must also be true?

Answer

We know that the expression must be negative. Therefore one or all of the terms x7, y8 and z10 must be negative; however, even powers always produce positive numbers, so y8 and z10 will both be positive. Odd powers can produce both negative and positive numbers, depending on whether the base term is negative or positive. In this case, x7 must be negative, so x must be negative. Thus, the answer is x < 0.

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Question

If p and q are positive integrers and 27p = 9q, then what is the value of q in terms of p?

Answer

The first step is to express both sides of the equation with equal bases, in this case 3. The equation becomes 33p = 32q. So then 3p = 2q, and q = (3/2)p is our answer.

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Question

Simplify 272/3.

Answer

272/3 is 27 squared and cube-rooted. We want to pick the easier operation first. Here that is the cube root. To see that, try both operations.

272/3 = (272)1/3 = 7291/3 OR

272/3 = (271/3)2 = 32

Obviously 32 is much easier. Either 32 or 7291/3 will give us the correct answer of 9, but with 32 it is readily apparent.

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Question

If and are integers and

$\left ( \frac{1}{3} \right )^{a}=27^{b}$

what is the value of $a\div b$ ?

Answer

To solve this problem, we will have to take the log of both sides to bring down our exponents. By doing this, we will get $\dpi{100} \small a\ast log\left (\frac{1}{3} \right )= b\ast log\left ( 27 \right )$ .

To solve for $\dpi{100} \small \frac{a}{b}$ we will have to divide both sides of our equation by $\dpi{100} \small log\frac{1}{3}$ to get $\dpi{100} \small \frac{a}{b}=\frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}$ .

$\dpi{100} \small \frac{log\left ( 27 \right )}{log\left ( \frac{1}{3} \right )}$ will give you the answer of –3.

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Question

If $\log 2=0.301$ and $\log 3=0.477$ , then what is $\log 12$ ?

Answer

We use two properties of logarithms:

$log(xy) = log (x) + log (y)$

$log(x^{n}) = nlog (x)$

So $\log 12=2 \log2+\log3$

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Question

Evaluate:

$x^{-3}x^{6}$

Answer

$x^{m}\ast x^{n} = x^{m + n}$ , here and , hence .

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Question

Solve for

$\left ( \frac{2}{3} \right )^{x+1} = \frac{27}{8}$

Answer

$\left ( \frac{2}{3} \right )^{x+1} = \frac{27}{8}$ = $\left ( \frac{3}{2} \right )^{3} = \left ( \frac{2}{3} \right )^{-3}$

which means

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Question

Which of the following statements is the same as:

$\dfrac{4^{x+2}}{2^y}$

$I)\ 2^{2x+4-y} \quad II)\ 2^{x+2} / 4^y \quad III)\ (4^x)( 4^2 )(2^{-y} )$

Answer

Remember the laws of exponents. In particular, when the base is nonzero:

$b^{m+n} = b^m b^n \qquad (b^m)^n = b^{mn} \ (b \times c)^n = b^n c^n \qquad b^{-n} = 1/b^n$

An effective way to compare these statements, is to convert them all into exponents with base 2. The original statement becomes:

$4^{x+2} / 2^y = 4^{x+2} (2^{-y}) = (2^2)^{x+2} 2^{-y} = 2^{2(x+2) } 2^{-y} = 2^{2x+4-y}$

This is identical to statement I. Now consider statement II:

$2^{x+2} / 4^y = 2^{x+2} ( 4^{-y} ) = 2^{x+2} ( 2^2 )^{-y} = 2^{x+2} 2^{-2y} = 2^{x+2-2y} eq 2^{2x+4-y}$

Therefore, statement II is not identical to the original statement. Finally, consider statement III:

$(4^x)(4^2 ) (2^{-y}) = (2^2)^x (2^2)^2 (2^{-y}) = (2^{2x} )( 2^4 ) (2^{-y}) = 2^{2x+4-y}$

which is also identical to the original statement. As a result, only I and III are the same as the original statement.

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Question

Write in exponential form:

$\sqrt[3]{\left ( x-1 \right )^{2}}$

Answer

Using properties of radicals e.g., $\sqrt[m]{x^{n}} = x^{\frac{n}{m}}$

we get $\sqrt[3]{\left ( x-1 \right )^{2}} = \left ( x-1 \right )^{\frac{2}{3}}$

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Question

Write in exponential form:

$\sqrt[3]{x^{5}}$

Answer

Properties of Radicals

$\sqrt[]{x} = x^{\frac{1}{2}}$

$\sqrt[3]{x} = x^{\frac{1}{3}}$

$x^{\frac{5}{3}}$

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Question

Write in radical notation:

$\left ( x+3 \right )^{\frac{2}{3}}$

Answer

Properties of Radicals

$\sqrt[]{x} = x^{\frac{1}{2}}$

$\sqrt[3]{x} = x^{\frac{1}{3}}$

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Question

Express in radical form : $x^{\frac{3}{5}}$

Answer

Properties of Radicals

$\sqrt[]{x} = x^{\frac{1}{2}}$

$\sqrt[3]{x} = x^{\frac{1}{3}}$

$x^{\frac{3}{5}}=\sqrt[5]{x^{3}}$

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Question

Simplify: $\sqrt[3]{2}\star \sqrt[3]{32}$

Answer

$\sqrt[3]{2}\star \sqrt[3]{32}= \sqrt[3]{64}= 4$

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Question

Simplify:

$\frac{\sqrt[3]{x^{5}y^{-2}}}{\sqrt[3]{x^{-1}y^{4}}}$

Answer

Convert the given expression into a single radical e.g. $\sqrt[3]{\frac{x^{5}y^{-2}}{x^{-1}y^{4}}}$ the expression inside the radical is:

$\frac{x^{6}}{y^{6}}$

and the cube root of this is :

$\frac{x^{2}}{y^{2}}$

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Question

Solve for .

$\log _{\frac{1}{5}}\left ( \frac{1}{25} \right )=x$

Answer

$\frac{1}{25}= \left ( \frac{1}{5} \right )^{x}$

Hence must be equal to 2.

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Question

Simplify:

$log\left (\frac{x}{y^{2}} \right )$

Answer

$log\left ( \frac{x}{y^{2}} \right )=logx-logy^{2}$

Now $logy^{2} = 2logy$

Hence the correct answer is

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Question

Solve for .

$\ln x=\ln (x-1)-\ln (x+3)$

Answer

If we combine into a single logarithmic function we get:

$\ln x=\ln \frac{x-1}{x+3}$

$x = \frac{x-1}{x+3}$

Solving for we get .

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