SAT 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.

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.

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 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 :

$4^{3t+ 8} = 8 ^{2t- 11}$

This statement has no solution.

$t = \frac{9}{2}$

Explanation

To solve this problem, first convert all numbers to have the like base of two

$4^{3t+ 8} = 8 ^{2t- 11}$

$\left (2^{2} \right )^{3t+ 8} = \left (2^{3} \right ) ^{2t- 11}$

Recall that when exponents are raised to another power the exponents are multiplied together.

$2^{2(3t+ 8)} = 2^{3(2t- 11)}$

Since we have the same base we can set the exponents equal to each other and solve.

This statement is identically false, so the original statement has no solution.

$3 ^{x}= 9 ^{y+1}$ .

Express in terms of .

$y = \frac{1}{2}x- 1$

$y = \frac{1}{2}x- \frac{1}{2}$

$y = x- \frac{1}{2}$

Explanation

$3 ^{x}= 9 ^{y+1}$

$3 ^{x}= (3 ^{2} )^{y+1}$

$3 ^{x}= 3 ^{2(y+1)} = 3 ^{2y+2}$

$y = \frac{x-2}{2} = \frac{1}{2}x- 1$

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

Solve for :

$3^{4y-2 } = 27 ^{y}$

$y = \frac{1}{2}$

$y =- \frac{1}{2}$

The equation has no solution.

Explanation

To solve this problem, first convert all numbers to have the like base of three.

$3^{4y-2 } = 27 ^{y}$

$3^{4y-2 } = \left ( 3^{3} \right ) ^{y}$

$3^{4y-2 } = 3^{3y}$

Since we have the same base we can set the exponents equal to each other and solve.

$16^{5x-17} = 8^{2y + 1}$

Express in terms of .

$y = \frac{10}{3}x- \frac{71}{6}$

$y = \frac{10}{3}x- \frac{65}{6}$

$y = \frac{10}{3}x- 10$

Explanation

$16^{5x-17} = 8^{2y + 1}$

$\left (2^{4} \right )^{5x-17} =( 2 ^{3})^{2y + 1}$

$2^{4(5x-17) }= 2 ^{3(2y + 1)}$

$2^{20x-68}= 2 ^{6y + 3}$

$\frac{6y}{6} = \frac{20x-71}{6}$

$y = \frac{10}{3}x- \frac{71}{6}$

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

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

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

How to find patterns in exponents

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