Which expression is equivalent to (7^3)^-2?
You multiply the powers.
\( \color{Black}{\Rightarrow (7^3)^{-2} = 7^{-6}}\)
ok the answers we can choose from are 7 1/7 1 over six 7's or 1 over seven 6's
\[\large (a^m)^n = a^{mn}\]
ok i get 7^6 but how do i get it to the correct answer 7 1/7 1 over six 7's or 1 over seven 6's
okay, one more thing, \[\large a^{-b}=\frac 1 {a^b}\] so, what's \(7^{-6}\) then?
@brett183 ??
i am thinking it is the 1 over the six 7's but not sure
\[\frac{1}{7}^6\]
yeah, you're thinking the right way, correct-o!! @brett183
@TheViper Uh-oh slight mistake in your post
yeah
what
oh never mind
so i multiply the powers then put a one over the 7 with the power of 6 and that tells me how many 7's i need right
exactly!
thank you
i have another question do you have time
Explain how you would find the product of 20^12 and 20^8. Include either the exponent rule or the expanded method in your explanation. Give your answer in exponential form.
Hmm, use this funda: \[\large a^m \times a^n = a^{m+n}\] Can you use this?
i dont know it is a esay question. and i am taking a online class.
yeah so use this formula, the 'a' in your case is 20, the 'm' is 12, and the 'n' is '8', so, \(20^{12} \times 20^8 = ??\) just use the formula i posted above
ok so would i do the same for this one. A kilobyte is 210. A megabyte is 220 bytes. How many kilobytes are in a megabyte? Use the exponent rule for division to find your answer and give your answer in exponential form.
but subtract
you mean \(2^{10}\) and \(2^{20}\) right? yeah, so you would divide, and this time, we subtract the exponents! \[\large \frac{a^m}{a^n}= a^{m-n}\]
yeap thats what i mean. thank you
Hey please post new questions in fresh threads after 'closing' this one. Thanks :)
hey i have one that i got wrong but i didnt ask you this one. simplify 15^-4 over 15^5. i answered 15^1
sorry didnt see your post. before i sent the question
umm, when it's "over", means you're dividing, so you subtract the exponents. so, \[\large \frac{15^{-4}}{15^5} = 15^{-5-4} = 15^{-9}\]
lol no worries ;)
ok thank u
anytime!
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