How to add exponents - PSAT Math

_a_4 = _a_2 * _a_2 and _b_6= _b_2 * _b_2 * _b_2

Therefore _a_4 + _b_6 = 35 * 35 + 52 * 52 * 52 = 1,225 + 140,608 = 141,833

Since we have two ’s in we will need to combine the two terms.

For this can be rewritten as

So we have .

Or

Divide this by :

Thus or

*Hint: If you are really unsure, you could have plugged in the numbers and found that the first choice worked in the equation.

The answer is 5.

8 + 2x+1 = 72

2x+1 = 64

2x+1 = 26

x + 1 = 6

x = 5

We can solve by converting all terms to a base of two. 4, 16, and 32 can all be expressed in terms of 2 to a standard exponent value.

We can rewrite the original equation in these terms.

Simplify exponents.

Finally, combine terms.

From this equation, we can see that .

We want to simplify 5_b_ – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) .

Notice that we can collect the –5(b–1) terms, because they are like terms. There are 5 of them, so that means we can write –5(b–1) – 5(b–1) – 5(b–1) – 5(b–1) – 5(b–1) as (–5(b–1))5.

To summarize thus far:

5_b_ – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–_1) – 5(_b–1) = 5_b +(–5(_b–_1))5

It's important to interpret –5(b–1) as (–1)5(b–1) because the –1 is not raised to the (b – 1) power along with the five. This means we can rewrite the expression as follows:

5_b_ +(–5(b–1))5 = 5_b_ + (–1)(5(b–1))(5) = 5_b_ – (5(b–1))(5)

Notice that 5(b–1) and 5 both have a base of 5. This means we can apply the property of exponents which states that, in general, abac = a b+c. We can rewrite 5 as 51 and then apply this rule.

5_b_ – (5(_b–1))(5) = 5_b – (5(_b–1))(51) = 5_b – 5(_b–_1+1)

Now, we will simplify the exponent b – 1 + 1 and write it as simply b.

5_b_ – 5(b–1+1) = 5_b – 5_b = 0

The answer is 0.

25left ( $5^{r}$ right ) is equal to which is equal to left ( $5^{r+2}$ right ). If we compare this to the original equation we get r+2=s-2rightarrow s=r+4

Combining the powers, we get $1024=2^{x}$.

From here we can use logarithms, or simply guess and check to get x=10.

Step 1: left ( $x^{-2}$$x^{5}$ right )= $x^{3}$

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

Step 3: (Correct Answer): $$\frac{x^{3}$$}{y}

Step 1: $$\frac{y^{5}$$}{left ( $x^{3}$$x^{2}$ right )left right $)y^{-1}$}

Step 2: $\frac{left ( $y^{5}$$$y^{1}$ right $)}{x^{3}$$x^{2}$}

Step $3:$\frac{y^{6}$$$}{x^{5}$}

Step 1:$\frac{1}{left ( -11 right $)^{12}$$left ( -11 right $)^{8}$}

Step 2: The above is equal to $\frac{1}{left ( -11 right $)^{20}$$}

-left ( -3 right $)^{0}$= -1

Similarly

So

A number in base five has powers of five as its place values; starting at the right, they are

The lowest base five number with four digits would be

in base ten.

The lowest base five number with five digits would be

in base ten.

Therefore, a number that is expressed as a four-digit number in base five must fall in the range

Three of the four numbers - all except 100 - fall in this range.

Using exponents, 27 is equal to 33. So, the equation can be rewritten:

34_x_ + 6 = (33)2_x_

34_x_ + 6 = 36_x_

When both side of an equation have the same base, the exponents must be equal. Thus:

4_x_ + 6 = 6_x_

6 = 2_x_

x = 3

_a_4 = _a_2 * _a_2 and _b_6= _b_2 * _b_2 * _b_2

Therefore _a_4 + _b_6 = 35 * 35 + 52 * 52 * 52 = 1,225 + 140,608 = 141,833

Since we have two ’s in we will need to combine the two terms.

For this can be rewritten as

So we have .

Or

Divide this by :

Thus or

*Hint: If you are really unsure, you could have plugged in the numbers and found that the first choice worked in the equation.

The answer is 5.

8 + 2x+1 = 72

2x+1 = 64

2x+1 = 26

x + 1 = 6

x = 5

We can solve by converting all terms to a base of two. 4, 16, and 32 can all be expressed in terms of 2 to a standard exponent value.

We can rewrite the original equation in these terms.

Simplify exponents.

Finally, combine terms.

From this equation, we can see that .