Calculate Expressions With Exponents: Step-by-Step Guide

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Hey guys! Let's break down these exponent calculations together. We'll go through each problem step by step, making sure you understand exactly how to solve them. Whether you're tackling algebra homework or just brushing up on your math skills, this guide will help you master these types of expressions.

1) Calculating 0.54β‹…240.5^{4} \cdot 2^{4}

When you're dealing with exponents, remember that a number raised to a power means multiplying the number by itself that many times. So, 0.540.5^{4} means 0.5 multiplied by itself four times, and 242^{4} means 2 multiplied by itself four times. Let's dive in! In this first expression, we need to calculate 0.54β‹…240.5^{4} \cdot 2^{4}. The key here is to recognize a property of exponents: when you have the same exponent applied to different bases, you can multiply the bases first and then apply the exponent.

Why does this work? Well, it's because (aβ‹…b)n=anβ‹…bn(a \cdot b)^{n} = a^{n} \cdot b^{n}. So, let's rewrite the expression:

0.54β‹…24=(0.5β‹…2)40. 5^{4} \cdot 2^{4} = (0.5 \cdot 2)^{4}

Now, 0.5 multiplied by 2 is simply 1. So, we have:

(0.5β‹…2)4=14(0.5 \cdot 2)^{4} = 1^{4}

And 1 raised to any power is just 1. So:

14=11^{4} = 1

Therefore, 0.54β‹…24=10.5^{4} \cdot 2^{4} = 1. See? That wasn't so bad!

2) Calculating (βˆ’0.125)3β‹…(βˆ’8)3(-0.125)^{3} \cdot (-8)^{3}

Next up, we have (βˆ’0.125)3β‹…(βˆ’8)3(-0.125)^{3} \cdot (-8)^{3}. This one might look a bit trickier with the negative signs and decimals, but don't worry, we'll handle it the same way. Again, we can use the property (aβ‹…b)n=anβ‹…bn(a \cdot b)^{n} = a^{n} \cdot b^{n}.

So, let's rewrite the expression:

(βˆ’0.125)3β‹…(βˆ’8)3=(βˆ’0.125β‹…βˆ’8)3(-0.125)^{3} \cdot (-8)^{3} = (-0.125 \cdot -8)^{3}

Now, we need to multiply -0.125 by -8. Remember, a negative times a negative is a positive. If you recognize that 0.125 is 1/8, then you can see that:

βˆ’0.125β‹…βˆ’8=βˆ’18β‹…βˆ’8=1-0.125 \cdot -8 = \frac{-1}{8} \cdot -8 = 1

So our expression becomes:

(βˆ’0.125β‹…βˆ’8)3=13(-0.125 \cdot -8)^{3} = 1^{3}

And just like before, 1 raised to any power is 1:

13=11^{3} = 1

Therefore, (βˆ’0.125)3β‹…(βˆ’8)3=1(-0.125)^{3} \cdot (-8)^{3} = 1. We're on a roll!

3) Calculating (32)10β‹…(113)10(\frac{3}{2})^{10} \cdot (1\frac{1}{3})^{10}

Alright, let's tackle (32)10β‹…(113)10(\frac{3}{2})^{10} \cdot (1\frac{1}{3})^{10}. This one involves fractions, but we'll use the same exponent property. First, we need to convert the mixed number 1131\frac{1}{3} into an improper fraction. To do this, we multiply the whole number (1) by the denominator (3) and add the numerator (1), then put it over the original denominator:

113=(1β‹…3)+13=431\frac{1}{3} = \frac{(1 \cdot 3) + 1}{3} = \frac{4}{3}

Now we can rewrite the original expression:

(32)10β‹…(113)10=(32)10β‹…(43)10(\frac{3}{2})^{10} \cdot (1\frac{1}{3})^{10} = (\frac{3}{2})^{10} \cdot (\frac{4}{3})^{10}

Using the property (aβ‹…b)n=anβ‹…bn(a \cdot b)^{n} = a^{n} \cdot b^{n} again:

(32)10β‹…(43)10=(32β‹…43)10(\frac{3}{2})^{10} \cdot (\frac{4}{3})^{10} = (\frac{3}{2} \cdot \frac{4}{3})^{10}

Now we multiply the fractions. Notice that we can simplify before multiplying by canceling out the 3s:

32β‹…43=32β‹…43=42\frac{3}{2} \cdot \frac{4}{3} = \frac{\cancel{3}}{2} \cdot \frac{4}{\cancel{3}} = \frac{4}{2}

And 42\frac{4}{2} simplifies to 2. So we have:

(32β‹…43)10=210(\frac{3}{2} \cdot \frac{4}{3})^{10} = 2^{10}

Now we just need to calculate 2102^{10}. This means 2 multiplied by itself 10 times:

210=2β‹…2β‹…2β‹…2β‹…2β‹…2β‹…2β‹…2β‹…2β‹…2=10242^{10} = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 1024

Therefore, (32)10β‹…(113)10=1024(\frac{3}{2})^{10} \cdot (1\frac{1}{3})^{10} = 1024. Awesome!

4) Calculating (βˆ’827)6β‹…(βˆ’38)6(\frac{-8}{27})^{6} \cdot (\frac{-3}{8})^{6}

Next, let's look at (βˆ’827)6β‹…(βˆ’38)6(\frac{-8}{27})^{6} \cdot (\frac{-3}{8})^{6}. This one involves negative fractions and a higher exponent, but the principles remain the same. We'll use that trusty property (aβ‹…b)n=anβ‹…bn(a \cdot b)^{n} = a^{n} \cdot b^{n} one more time:

(βˆ’827)6β‹…(βˆ’38)6=(βˆ’827β‹…βˆ’38)6(\frac{-8}{27})^{6} \cdot (\frac{-3}{8})^{6} = (\frac{-8}{27} \cdot \frac{-3}{8})^{6}

Now, let's multiply the fractions. Again, we can simplify before multiplying:

βˆ’827β‹…βˆ’38=βˆ’827β‹…βˆ’38=19\frac{-8}{27} \cdot \frac{-3}{8} = \frac{-\cancel{8}}{27} \cdot \frac{-3}{\cancel{8}} = \frac{1}{9}

Why 19\frac{1}{9}? Well, the two negatives cancel each other out, making the result positive. And if you divide 27 by 3 you get 9. Remember 3 divided by 3 is 1.

So our expression becomes:

(βˆ’827β‹…βˆ’38)6=(19)6(\frac{-8}{27} \cdot \frac{-3}{8})^{6} = (\frac{1}{9})^{6}

Now we need to calculate (19)6(\frac{1}{9})^{6}. Remember that (19)6(\frac{1}{9})^{6} means 1696\frac{1^{6}}{9^{6}}. 161^{6} is just 1. And 969^{6} is a bit more involved, but let's think about it. 9 is 323^{2}, so 969^{6} is (32)6(3^{2})^{6}, which is 3123^{12} (using the power of a power rule, (am)n=amβ‹…n(a^{m})^{n} = a^{m \cdot n}):

(19)6=196=1(32)6=1312(\frac{1}{9})^{6} = \frac{1}{9^{6}} = \frac{1}{(3^{2})^{6}} = \frac{1}{3^{12}}

To find 3123^{12}, you'd multiply 3 by itself 12 times, which gives you 531441. So:

1312=1531441\frac{1}{3^{12}} = \frac{1}{531441}

Therefore, (βˆ’827)6β‹…(βˆ’38)6=1531441(\frac{-8}{27})^{6} \cdot (\frac{-3}{8})^{6} = \frac{1}{531441}. Phew, that was a bigger number!

5) Calculating 35β‹…2665\frac{3^{5} \cdot 2^{6}}{6^{5}}

Let's jump into 35β‹…2665\frac{3^{5} \cdot 2^{6}}{6^{5}}. This one involves division and different exponents, but we'll break it down. First, we need to recognize that 6 can be written as 2β‹…32 \cdot 3. So, 656^{5} can be rewritten as (2β‹…3)5(2 \cdot 3)^{5}. Using the power of a product rule, (aβ‹…b)n=anβ‹…bn(a \cdot b)^{n} = a^{n} \cdot b^{n}, we get:

65=(2β‹…3)5=25β‹…356^{5} = (2 \cdot 3)^{5} = 2^{5} \cdot 3^{5}

Now we can rewrite the original expression:

35β‹…2665=35β‹…2625β‹…35\frac{3^{5} \cdot 2^{6}}{6^{5}} = \frac{3^{5} \cdot 2^{6}}{2^{5} \cdot 3^{5}}

Now we can simplify by canceling out common factors. We have 353^{5} in both the numerator and the denominator, so they cancel out. We also have 262^{6} in the numerator and 252^{5} in the denominator. When dividing exponents with the same base, you subtract the exponents:

2625=26βˆ’5=21=2\frac{2^{6}}{2^{5}} = 2^{6-5} = 2^{1} = 2

So, our simplified expression is:

35β‹…2625β‹…35=2\frac{3^{5} \cdot 2^{6}}{2^{5} \cdot 3^{5}} = 2

Therefore, 35β‹…2665=2\frac{3^{5} \cdot 2^{6}}{6^{5}} = 2. Nice and tidy!

6) Calculating 38β‹…2666\frac{3^{8} \cdot 2^{6}}{6^{6}}

Last but not least, we have 38β‹…2666\frac{3^{8} \cdot 2^{6}}{6^{6}}. We'll use the same strategy as before. We know that 6=2β‹…36 = 2 \cdot 3, so 66=(2β‹…3)66^{6} = (2 \cdot 3)^{6}. Using the power of a product rule:

66=(2β‹…3)6=26β‹…366^{6} = (2 \cdot 3)^{6} = 2^{6} \cdot 3^{6}

Now we rewrite the original expression:

38β‹…2666=38β‹…2626β‹…36\frac{3^{8} \cdot 2^{6}}{6^{6}} = \frac{3^{8} \cdot 2^{6}}{2^{6} \cdot 3^{6}}

Again, we can simplify by canceling out common factors. This time, 262^{6} appears in both the numerator and the denominator, so they cancel out. We also have 383^{8} in the numerator and 363^{6} in the denominator. When dividing exponents with the same base, you subtract the exponents:

3836=38βˆ’6=32\frac{3^{8}}{3^{6}} = 3^{8-6} = 3^{2}

So, our simplified expression is:

38β‹…2626β‹…36=32\frac{3^{8} \cdot 2^{6}}{2^{6} \cdot 3^{6}} = 3^{2}

And 323^{2} is simply 3 multiplied by itself, which is 9:

32=3β‹…3=93^{2} = 3 \cdot 3 = 9

Therefore, 38β‹…2666=9\frac{3^{8} \cdot 2^{6}}{6^{6}} = 9. We did it!

Conclusion

So, guys, we've walked through six different exponent calculations, using key properties and breaking down each problem step by step. Remember, the key is to recognize the properties of exponents and simplify wherever possible. Keep practicing, and you'll become an exponent expert in no time!