PSAT Math - How to find patterns in exponents

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

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

Explanation

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.

Simplify 272/3.

125

729

Explanation

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.

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

(2/3)p

(3/2)p

Explanation

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.

If and are integers and

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

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

$\frac{1}{3}$

$-\frac{1}{3}$

Explanation

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.

Solve for

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

None of the above

Explanation

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

which means

Write in radical notation:

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

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

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

$\sqrt[3]{\left ( x+3 \right )}$

$\sqrt[]{\left ( x+3 \right )}$

$x^{\frac{2}{3}} + 3^{\frac{2}{3}}$

Explanation

Properties of Radicals

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

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

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

Explanation

We use two properties of logarithms:

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

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

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

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

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

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

$x^{0.6}$

$x^{15}$

$\sqrt[3]{x}$

Explanation

Properties of Radicals

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

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

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

Write in exponential form:

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

$\left ( x-1 \right )^{\frac{2}{3}}$

$\left ( x-1 \right )$

$\left ( x-1 \right )^{\frac{3}{2}}$

$\left ( x-1 \right )^{2}$

$\left ( x-1 \right )^{3}$

Explanation

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

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

Which of the following statements is the same as:

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

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

$I \mbox{ and } III \mbox{ only}$

$I \mbox{ only}$

$I \mbox{ and } II \mbox{ only}$

$II \mbox{ and } III \mbox{ only}$

$I, II \mbox{ and } III$

Explanation

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.

How to find patterns in exponents

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