Posts tagged with: exponents

When you have exponents with the same base and you divide them, you subtract the exponents.

So from what we’re given, we know that 2x – (xy) = 1. We can solve that for y.

2x – (xy) = 1
2x – x + y = 1
x + y = 1
y = 1 – x

from Tumblr https://ift.tt/2PAzwqC

Practice test 8 Calculator #13

First, you can plug in on this one, so if you feel rusty on your exponent rules at all, that’s a good move. Especially on the calculator section. Say, for example, that you plug in 4 for a. Just enter it all into your calculator (you may need to be careful with parentheses in the exponent depending on the kind of calculator you have):

    \begin{align*}4^{-\frac{1}{2}}&=x\\0.5&=x\end{align*}

Now that you know x, plug 0.5 into each answer choice to see which one gives you 4.

A) \sqrt{0.5}\approx 0.707

B) -\sqrt{0.5}\approx -0.707

C) \frac{1}{0.5^2}=4

D) -\frac{1}{0.5^2}=-4

Obviously, C must be the answer.

To solve this algebraically, first start by squaring both sides. Raising a power to a power is the same as multiplying the powers, so that’ll get rid of the 1/2 on the left:

    \begin{align*}\left(a^{-\frac{1}{2}}\right)^2&=x^2\\a^{-1}&=x^2\end{align*}

Now raise both sides to the –1 power to get a truly alone. Remember that a negative exponent is the same as 1 over the positive exponent, so you can transform the right hand side from x^{-2} to \frac{1}{x^2} to finish the problem.

    \begin{align*}\left(a^{-1}\right)^{-1}&=\left(x^2\right)^{-1}\\a&=x^{-2}\\a&=\frac{1}{x^2}\end{align*}

\sqrt{x-k}=k

None. The square root of something will never be negative, so if k < 0, it can’t be all alone on the right side of that equation.

from Tumblr https://ift.tt/2Aah6qp

None. The square root of something will never be negative, so if k < 0, it can’t be all alone on the right side of that equation.

from Tumblr https://ift.tt/2Oaom8A

Question from March 2018 SAT: section 3 #15
Which of the following expressions is equivalent to (-4x^3)^2/3 ?

A. -2x^3 * ∛2
B. -x^3 * ∛16
C. 2x^2 * ∛2
D. 2x^2 * ∛16

I like this question. For fractional exponents, remember that the top number is the power and the bottom is the root. Also remember that when you’re applying powers to an expression like this, you have to apply the power to all parts being multiplied. So what I’m going to do first below is break the –4 off, and then apply the exponents.

    \begin{align*}&~~~\left(-4x^3\right)^\frac{2}{3}\\&=\left(-4\right)^\frac{2}{3}\left(x^3\right)^\frac{2}{3}\\&=\left(16\right)^\frac{1}{3}\left(x^6\right)^\frac{1}{3}\\&=\sqrt[3]{16}\left(x^2\right)\end{align*}

Unfortunately, that’s not an answer choice. It is good enough to eliminate choices A and B, though, because there’s nothing else we can do to the x term. It’s also good enough to eliminate D if we’re in a hurry because D makes a 2 magically appear without changing anything else in our simplified expression. Since we’re not in a rush, though, let’s manipulate what we’ve got into choice C just to be sure.

\begin{align*}&~~~~\sqrt[3]{16}\left(x^2\right)\\&=\sqrt[3]{(2)(2)(2)(2)}\left(x^2\right)\\&=\sqrt[3]{\left(2^3\right)(2)}\left(x^2\right)\\&=\sqrt[3]{\left(2^3\right)}\sqrt[3]{2}\left(x^2\right)\\&=2x^2\cdot \sqrt[3]{2}\end{ailgn*}

This no-calc prompt is from the Daily Practice app:

The temp T in °C of a chilled drink after the drink has been sitting on a table for m min is represented by T(m) = 32 – 28*3^(-0.05m).

What is the best interpretation of the # 32 in this context?

A. The drink is originally 32°C.
B. Every 32 min, the temperature warms by 3°C.
C. After 32 min, the drink will fully warm to the ambient temp.
D. After the drink has been sitting for a very long time, the temp of the drink will approach 32°C.

I’m going to try to talk you through my thought process for this question rather than just do math.

First step for me here is to eliminate B and C—we know that m is minutes and that appears in the exponent in this function, far away from the 32.

From there,* your job is to figure out whether the expression will approach 32 as m increases (in which case the answer is D) or get farther away from 32 as m increases (in which case the answer is A).

Think about the equation this way: T(m)=32-\text{[something]}. Of course, that something is 28\times 3^{-0.05m}, but stay with me for a minute. If the [something] is going to grow, then A will be the answer. If the something is going to shrink, then D will be the answer.

Now, does 28\times 3^{-0.05m} grow or shrink?

We know a negative exponent is the same as the inverse of the positive exponent, so let’s look at it that way:

    \begin{align*}28\times \frac{1}{3^{0.05m}}\end{align*}

When m = 0, that’s going to just be 28:

    \begin{align*}28\times \frac{1}{3^{0.05(0)}}\\28\times \frac{1}{3^0}}\\28\times \frac{1}{1}\\28\end{align*}

When m grows to 20, however…

    \begin{align*}28\times \frac{1}{3^{0.05(20)}}\\28\times \frac{1}{3^1}}\\28\times \frac{1}{3}\\9.333...\end{align*}

When m grows, our [something] shrinks! Therefore, the function will get closer to 32 as m increases, and D is the answer!

 

* From the get-go, you might also be able to eliminate A if you know a few things about Celsius. 32° C is about 90° F, which is in no way a “chilled” drink. In that case, you really need not do any math at all!

How do you solve Practice test 6 Section 3 Number 16?

Remember that, for questions like this one, you only need to find one solution, not all solutions. Therefore, go for the simplest one you can!

Think of powers you know that land on 16. The first one that comes to mind for me is 4^2=16. The only restriction the question places on a and b is that they must both be positive integers, so let’s pick numbers that will make the given equation the same as 4^2=16.

Obviously we can make the base a = 4. How do we make the exponent equal to 2? Well, \dfrac{8}{4}=2, so let’s say b = 8.

    \begin{align*}a^\frac{b}{4}&=16\\\\4^\frac{8}{4}&=16\\\\4^2&=16\end{align*}

So there you go: 8 is one valid answer. For each of the other answers (1, 2, 4, 8, and 16 are all valid) there’s an exponential expression that works. While I find some of the following routes a little less obvious, all are valid:

    \begin{align*}2^4=16\qquad\Longrightarrow\qquad b=16\end{align*}

    \begin{align*}16^1=16\qquad\Longrightarrow\qquad b=4\end{align*}

    \begin{align*}256^\frac{1}{2}=16\qquad\Longrightarrow\qquad b=2\end{align*}

    \begin{align*}65536^\frac{1}{4}=16\qquad\Longrightarrow\qquad b=1\end{align*}

Test 2 Section 3 #7

This is one of my favorite questions because there are a couple fun paths to take: one with pure exponent rules, and one with exponent rules and difference of two squares. So fun! (I understand if you don’t share my enthusiasm.)

First, the exponent-rules-only method. We need to take what we’re given and simplify it enough that we can stop worrying about x and focus exclusively on a and b. To do that, we first need to remember that when we raise a power to a power, we multiply the powers. In other words:

    \begin{align*}\dfrac{x^{a^2}}{x^{b^2}}&=x^{16}\\\\\dfrac{x^{2a}}{x^{2b}}&=x^{16}\end{align*}

From there, we need to remember that when we divide exponential expressions with the same base, we subtract the exponents.

    \begin{align*}\dfrac{x^{2a}}{x^{2b}}&=x^{16}\\\\x^{2a-2b}&=x^{16}\end{align*}

From there, we’re good to eliminate x and just say that 2a-2b=16. Since the question asks us for the value of a-b, just divide that equation by 2 to get a-b=8.

Going the second way involves the division exponent rule but not the power-to-a-power rule. Skip right to the division rule:

    \begin{align*}\dfrac{x^{a^2}}{x^{b^2}}&=x^{16}\\\\x^{a^2-b^2}&=x^{16}\end{align*}

From there, we can say that a^2-b^2=16 and use our difference of two squares skills.

    \begin{align*}a^2-b^2&=16\\(a+b)(a-b)&=16\end{align*}

The question tells us that a+b=2, so let’s substitute:

    \begin{align*}(a+b)(a-b)&=16\\2(a-b)&=16\\a-b&=8\end{align*}

So there you go—two fun ways to get this one done. Which do you prefer?

If r>0 and (9r/2)^(1/3) = (1/2) r, what is the value of r?

Two good ways to go here. First, the algebra:

    \begin{align*}\left(\dfrac{9r}{2}\right)^\dfrac{1}{3}&=\dfrac{1}{2}r\\\\\left(\left(\dfrac{9r}{2}\right)^\dfrac{1}{3}\right)^3&=\left(\dfrac{1}{2}r\right)^3\\\\\dfrac{9r}{2}&=\dfrac{1}{8}r^3\\\\36r&=r^3\end{align*}

One possible solution to the above is r = 0, but the question says that r > 0, so we can ignore that and move on:

    \begin{align*}36r&=r^3\\36&=r^2\\6&=r\end{align*}

We need not consider r = –6 because we know r is positive.

The other way to go is to graph both sides of the equation and find the intersection(s).

Because r is positive (my software made me use x in the graph above, as your calculator probably will, so just remember that xr), the only intersection that matters is (6, 3). At that intersection, x = 6, so we know that the value of r is 6.

would you please explain #12 in test 5, section 3.

 

Yeah, this one is trickier than it looks. I think most people get turned around on it because they try to get out of fractional exponent notation too early, and then they get stuck. It’s a LOT easier, in my opinion, to do a little manipulation in exponent form before you jump to radicals. In this case, that means making a 9=3^2 substitution, which will allow you to reduce the fraction.

    \begin{align*}&9^{3/4}\\=&\left(3^2\right)^{3/4}\end{align*}

Remember than when you raise a power to a power, you multiply those powers. So:

    \begin{align*}&9^{3/4}\\=&\left(3^2\right)^{3/4}\\=&3^{6/4}\\=&3^{3/2}\end{align*}

Now you can convert way more comfortably to radical notation. Remember, with fractional exponents, the numerator (top number in the fraction) is the power and the denominator (bottom number in the fraction) is the root.

    \begin{align*}&3^{3/2}\\=&\sqrt{3^3}\\=&\sqrt{3\times 3\times 3}\\=&3\sqrt{3}\end{align*}

Can you do Test 2 Section 3 #14? Thanks!

The thing to know here is that the basic form for exponential growth or decay is this:

\text{Value after }n\text{ periods}=(\text{Starting Value})(1+\text{Growth Rate per Period})^n

When this question tells you that the substance decays at an annual rate of 13%, that means you have a growth rate of –0.13. The question also tells you that the starting amount is 325 grams and that the period in years is represented by t.

We can plug the given information into the standard form:

f(t)=325(1+(-0.13))^t

Simplify the growth rate and you have your answer:

f(t)=325(0.87)^t

Can you do Test 6 section 4 number 37? I always miss this kind of question!

The best way to be sure you’re not missing the mark on a question like this is to actually list the years rather than just trying to capture everything in an equation. So say he starts with x dollars on January 1, 2001. You know his money doubles each year, so you can write the following.

January 1, 2001: x
January 1, 2002: 2x
January 1, 2003: 2(2x) = 4x
January 1, 2004: 2(4x) = 8x
January 1, 2005: 2(8x) = 16x

Of course, you know from the question that on January 1, 2005, Jeremy really had $480 in his account. Easy enough to solve from there:

16x = 480
x = 30

The unnecessarily complicated way to do this question is with an exponent formula. You know his money doubles every year for 4 years, so you can write:

480=x(2^4)

Of course, that just simplifies to the same equation we just solved: 16x = 480.

What is the fastest way to do test 5 #14 from the no calculator section?

One tricky bit of business in this question is that it gives you f(x) and then asks what the graph of -f(x) looks like. They’re not being as mean as they could be, though: none of the answer choices are what f(x) should look like, so even if you’re not reading carefully at first, you should be able to clear that first hurdle no problem. Anyway, if f(x)=2^x+1, then -f(x)=-2^x-1.

When you need to figure out what a graph should look like without the aid of your calculator, try to pick a few easy points to calculate and use those to eliminate choices. If you look at the answer choices here, for example, you’ll notice that most of them have different y-intercepts, so start by plugging x = 0 to find the y-intercept you seek.

-2^0-1

=-1-1 (Remember, anything to the 0 power is 1!)

=-2

Only choice C has a y-intercept of –2, so that must be the answer!

Test 1 Section 3 Number 20

This is an easy one to get tied up on! First, double the first equation so that you’ve got 2a= something in both equations.

    \begin{align*}2a&=10\sqrt{2}\\2a&=\sqrt{2x}\end{align*}

Now recognize that the question asks you nothing about a. It only wants you to solve for x! So use the transitive property to set up one equation with one variable:

    \begin{align*}10\sqrt{2}&=\sqrt{2x}\end{align*}

Solve by squaring both sides and then doing a little more algebraic manipulation:

    \begin{align*}\left(10\sqrt{2}\right)^2&=\left(\sqrt{2x}\right)^2\\(10^2)(2)&=2x\\200&=2x\\100&=x\end{align*}

Test 1 Sec 3 14 please

There are two ways to go here. First, exponent rules.

Recognize that 8=2^3, so you can rewrite \dfrac{8^x}{2^y} as \dfrac{\left(2^3\right)^x}{2^y}. Of course, that simplifies:

    \begin{align*}&\dfrac{\left(2^3\right)^x}{2^y}\\=&\dfrac{2^{3x}}{2^y}\\=&2^{3x-y}\end{align*}

Because you know that 3x-y=12, you know you’ve got 2^{12}.

The other way to go is to plug in: pick values for x and y such that 3x – y = 12. For example, you may pick x = 5 and y = 3, because 3(5) – 3 = 12. Then you can evaluate \dfrac{8^5}{2^3} and match it with an answer. However, since this is the no calculator section, you may decide that this is a suboptimal way to go because the numbers are a little too big to deal with easily unless you know your exponent rules (in which case you didn’t need to plug in in the first place).