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Thread: Simplifying expressions with fractional exponents and a variable in the numerator

  1. #1

    Question Simplifying expressions with fractional exponents and a variable in the numerator

    Hey guys, so this is my first post so bare with me. I am working a problem set in a precalc workbook and came across this:
    (3)(3x/2)2

    The answer i always arrive at no matter how I work it comes out 9^x but the answer sheet on the problem set says that it is 3^(x+1).
    Can someone give me a break down of how this is worked??? I've scoured everything but the dark web looking for a solution and can find nothing. Further more, I have found that I am having the same issue with every problem of this type. I feel I'm forgetting a simple rule here.

    fyi, I am teaching myself calculus as a hobby.

    Thanks in Advance,
    Joe
    Last edited by jtbez8809; 02-01-2018 at 01:26 PM.

  2. #2
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    Quote Originally Posted by jtbez8809 View Post
    Hey guys, so this is my first post so bare with me. I am working a problem set in a precalc workbook and came across this:
    (3)(3x/2)2

    The answer i always arrive at no matter how I work it comes out 9^x but the answer sheet on the problem set says that it is 3^(x+1).
    Can someone give me a break down of how this is worked??? I've scoured everything but the dark web looking for a solution and can find nothing. Further more, I have found that I am having the same issue with every problem of this type. I feel i'm forgetting a simple rule here.

    fyi, I am teaching myself calculus as a hobby.

    Thanks in Advanced,
    Joe
    You used the laws of exponents:

    (a^b)^c = a^(b*c)

    now you want to evaluate (3
    x/2)2

    Here

    a = 3

    b = x/2

    c = 2

    then

    [3^(x/2)]^2 = 3^[(x/2)*2] = 3^x

    then

    3 * [3^(x/2)]^2 = 3 * 3^[(x/2)*2] = 3 * 3^x = 3^(x+1)
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3

    ok...

    I got that far and when I get to the last computation in that line of simplification 3 * 3^x and I get 9^x...

    excuse my ignorance, where does the ^x+1 come from, specifically the + 1 part?

    I'm assuming that the terms are not like and you cannot multiply 3^x * 3????

    what am i missing, its driving me insane....

  4. #4
    Elite Member stapel's Avatar
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    Quote Originally Posted by jtbez8809 View Post
    I got that far and when I get to the last computation in that line of simplification 3 * 3^x and I get 9^x...
    How did you take (3^1)*(3^x), and get (3^2)^x? What exponent rule did you use?

  5. #5
    Quote Originally Posted by stapel View Post
    How did you take (3^1)*(3^x), and get (3^2)^x? What exponent rule did you use?
    ahhhhhhh.... the original question was to simplify and leave with out radicals. I used multiplying powers to find that (3)[3^(x/2)]^2 = (3)(3^x) ...for 2/1 * x/2 simplifies to x....leaves me with (3)(3^x). I'm assuming I'm making some simple stupid mistake here...I've gotten through the rest of the problem set and literally have tried this one 20 times over.

    HOW DO I GET TO 3^(X+1). I mean I can see (3)(3)^2/x) maybe = 3^x but where does the +1 come from....

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    Quote Originally Posted by jtbez8809 View Post
    ahhhhhhh.... the original question was to simplify and leave with out radicals. I used multiplying powers to find that (3)[3^(x/2)]^2 = (3)(3^x) ...for 2/1 * x/2 simplifies to x....leaves me with (3)(3^x). I'm assuming I'm making some simple stupid mistake here...I've gotten through the rest of the problem set and literally have tried this one 20 times over.

    HOW DO I GET TO 3^(X+1). I mean I can see (3)(3)^2/x) maybe = 3^x but where does the +1 come from....
    (3) * (3^x) = (3^1)*(3^x) = 3^(x+1)

    Are you working with pencil and paper - or staring at the screen?
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  7. #7
    Elite Member stapel's Avatar
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    Quote Originally Posted by stapel View Post
    How did you take (3^1)*(3^x), and get (3^2)^x? What exponent rule did you use?
    Quote Originally Posted by jtbez8809 View Post
    ...leaves me with (3)(3^x). I'm assuming I'm making some simple stupid mistake here...I've gotten through the rest of the problem set and literally have tried this one 20 times over.

    HOW DO I GET TO 3^(X+1). I mean I can see (3)(3)^2/x) maybe = 3^x but where does the +1 come from....
    Did you read what I asked? You had (3^1)*(3^x). How did you get to (3^2)^x? Where did you make the "1" disappear to?

    Please study the helps that you've been given, and reply showing your work, so we can try to figure out how you're still stuck. Thank you!

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    Elite Member stapel's Avatar
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    Received via private-messaging system; moved to tutoring forums for tutoring replies.

    Quote Originally Posted by jtbez8809 View Post
    If so, why do you not multiply the coefficients into 9^x+1?
    From whence are you getting a "9"?

    Quote Originally Posted by jtbez8809 View Post
    and further more, 3^1 should just be 1, so that would change the value of the equation and definitely not be simplification.
    How are you getting that anything to the power of 1 is "just 1"? (That would be "anything to the power of 0".)

    Quote Originally Posted by jtbez8809 View Post
    so i really couldn't follow your question. are you saying it should be ....

    ... = 3 ^1 * 3 ^x = 3^(x+1)
    Yes. To learn how exponents work, please try online lessons, such as are listed here. Then please show your steps and state your reasoning, so we can try to figure out what is going on. Thank you!
    Last edited by Subhotosh Khan; 02-03-2018 at 04:22 PM.

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    Empathy!

    Quote Originally Posted by jtbez8809 View Post
    Hey guys, so this is my first post so bare with me. I am working a problem set in a precalc workbook and came across this:
    (3)(3x/2)2

    The answer i always arrive at no matter how I work it comes out 9^x but the answer sheet on the problem set says that it is 3^(x+1).
    Can someone give me a break down of how this is worked??? I've scoured everything but the dark web looking for a solution and can find nothing. Further more, I have found that I am having the same issue with every problem of this type. I feel I'm forgetting a simple rule here.

    fyi, I am teaching myself calculus as a hobby.

    Thanks in Advance,
    Joe

    I've scoured everything but the dark web looking for a solution - ha ha ha I know the feeling-I'm teaching myself at present and when you get stuck for ages it can feel like **** until clarity dawns. I got stuck like you -keep going!

  10. #10

    Talking Help much appreciated

    I did mistakenly say that 3^1 should be 1. I've tried it multiple times on paper and I couldn't find the mistake I made. High school was 11 years ago guys and I have to work the problems to understand the rules. Now I got my shame message out of the way. I think you guys answered my question well with these questions....

    (3) * [(3^X/2)]^2 = (3) * [(3)^(X/2 * 2/1)] (raising a power to a power) = (3)(3^x) (to make 3 a like term and simplify further you raise it to the first power) then multiplying powers produces this

    3 ^x + 1

    I totally botched this by not realizing 1: 3^1 = 3 and 2: the rules for variables in exponent form may be the same but without knowing the value it is treated different in simplifying.

    Thanks a million guys.
    Sure this isn't the last you'll hear from me.

    Long road to working out Calc and QM

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