yes when you have a power times a power, you add the exponents so I think you are correct.
Hello everyone, I completed my last course in college algebra some 3 years ago, needless to say I am rusty at this point. Now I am close to graduating; however, I have learned that one of the requirements for graduate school is a full year of calculus. With that being said I have bought quit a few math related books, starting out with Algebra and working my way up. This is so I can eventually be prepared for calculus before ever stepping foot in the class. I also apologize as I could only find superscripts for 2 and 3 in word 2002.
With that being established currently in my book I’m working with algebraic operations, and I think the book is incorrect regarding a specific example.
Straight from the book on how to solve it:
4a² b³ (2a³b²) + 5ab^-2 (2a^4 b^7) +5
4 · 2a²a³b³b²+5 · 2aa^4 b^-2 b^7 + 5
- Multiply the variables together separately in each term
8a^5 b^5 + 10a^5 b^5 + 5
- Add the exponents of the variables that are alike
= (8 + 10) a^5 b^5 + 5 = 18a^5 b^5 + 5 for your final answer
- Combine terms that are alike
Making the transition from step 2 to 3 where did the other b^5 go? Since there are two b^5 shouldn’t the final answer be 18a^5 b^10 +5… it seems the author left a b^5. Would some one please confirm if this is correct, If I am wrong and the author is right would some one please instruct me as to where the other b^5 went?
Many thanks
Straight from the book on how to solve it:
4a² b³ (2a³b²) + 5ab^-2 (2a^4 b^7) +5
4 · 2a²a³b³b²+5 · 2aa^4 b^-2 b^7 + 5
- Multiply the variables together separately in each term
8a^5 b^5 + 10a^5 b^5 + 5 There are 3 terms here. Two of them you can combine.
- Add the exponents of the variables that are alike
combine into
Then just tack on the 5 and you have:
= (8 + 10) a^5 b^5 + 5 = 18a^5 b^5 + 5 for your final answer
- Combine terms that are alike
Making the transition from step 2 to 3 where did the other b^5 go? Since there are two b^5 shouldn’t the final answer be 18a^5 b^10 +5… it seems the author left a b^5. Would some one please confirm if this is correct, If I am wrong and the author is right would some one please instruct me as to where the other b^5 went?