Manipulating exponents: not sure how this example is worked out

• geologist
In summary, the problem on Brilliant.org may not be clear due to the omission of a step in solving 2^3 + 6^3.
geologist
Member has been advised to use the template and show some effort.
Summary: An example problem on Brilliant.org leaves out a step in solving 2^3 + 6^3

I've spend a couple days trying to understand how the second step of solving the problem 23 + 63, which is 23 + 63 = 23 + (23 * 33) = 23(1 + 33).

It is not clear how this problem gets to 23(1 + 33). I can see that somehow you have to factor one of the 23, but I cannot visualize how that is done. What concept should I review to understand this problem?

Just a quick look at your post #1,
...
DISTRIBUTIVE PROPERTY

jedishrfu
2^3+6^3
2^3+(2*3)^3
2^3+(2^3*3^3)
2^3+(2^3)(3^3)
(2^3)(1+3^3), factorizable because of the Distributive Property

jedishrfu
geologist said:
I cannot visualize how that is done.
It is how an area can be calculated:

jedishrfu
fresh_42, I have occasionally used that kind of visualizing, too. When trying to factorize an expression, one must actually see the factors IN the expression.

symbolipoint said:
2^3+6^3
2^3+(2*3)^3
2^3+(2^3*3^3)
2^3+(2^3)(3^3)
(2^3)(1+3^3), factorizable because of the Distributive Property

I understand how a(b-c) = ab - ac and how (5x+2)(5x+2) = 25x^2 + 20x + 4, but I'm struggling to understand what is "a" in the above problem. I'll have to digest this information and read more examples on the distributive property.

geologist said:
I understand how a(b-c) = ab - ac and how (5x+2)(5x+2) = 25x^2 + 20x + 4, but I'm struggling to understand what is "a" in the above problem. I'll have to digest this information and read more examples on the distributive property.
The expression you have is ##a+(a\cdot b)##. Now we first write it as ##a\cdot 1 + a\cdot b## so we have a common factor ##a## in both terms, which allows us to use the distributive law backwards:
$$a\cdot 1 + a\cdot b = a\cdot (1+b)$$

fresh_42 said:
The expression you have is ##a+(a\cdot b)##. Now we first write it as ##a\cdot 1 + a\cdot b## so we have a common factor ##a## in both terms, which allows us to use the distributive law backwards:
$$a\cdot 1 + a\cdot b = a\cdot (1+b)$$
That make sense. I understand what’s going on now. The step 2^3+(2^3*3^3) is unnecessary. My confusion with 2^3(1+3^3) can be understood better looking at the originally stated problem 2^3 + 3^3. Now it’s easy to see how 2^3 factors out.

geologist said:
That make sense. I understand what’s going on now. The step 2^3+(2^3*3^3) is unnecessary. My confusion with 2^3(1+3^3) can be understood better looking at the originally stated problem 2^3 + 3^3. Now it’s easy to see how 2^3 factors out.
I was trying to show 'all' the steps to make the solution clear. Maybe just a few minutes of rest, and returning to it , the steps will seem much more understandable than before. The way I showed association might have unsettled you when first looking.

What is an exponent?

An exponent is a number that represents how many times a base number should be multiplied by itself. It is written as a superscript number to the right of the base number.

How do you manipulate exponents?

To manipulate exponents, you can use the basic rules of exponents, such as the product rule, quotient rule, and power rule. You can also simplify expressions with exponents by combining like terms and using properties of exponents.

What is the order of operations for manipulating exponents?

The order of operations for manipulating exponents is to first simplify any expressions inside parentheses or brackets, then evaluate any exponents, and finally perform any multiplication or division from left to right.

Can exponents be negative or fractions?

Yes, exponents can be negative or fractions. A negative exponent indicates the reciprocal of the base raised to the positive exponent, while a fraction exponent indicates the root of the base raised to the numerator and denominator of the fraction.

Why is it important to understand how to manipulate exponents?

Understanding how to manipulate exponents is important in many fields of science, such as physics, chemistry, and engineering. It allows for simpler and more efficient calculations and can help solve complex problems involving exponential functions.

• Precalculus Mathematics Homework Help
Replies
13
Views
2K
• Programming and Computer Science
Replies
3
Views
706
• Quantum Physics
Replies
5
Views
997
• General Math
Replies
17
Views
1K
• Differential Equations
Replies
1
Views
2K
• Precalculus Mathematics Homework Help
Replies
7
Views
2K
• Precalculus Mathematics Homework Help
Replies
23
Views
1K
• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Engineering and Comp Sci Homework Help
Replies
9
Views
1K
• Precalculus Mathematics Homework Help
Replies
2
Views
1K