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anonymous

  • one year ago

Ciara solved the exponential equation 3x+1 = 15 and her work is shown below. What is the first step she did incorrectly? Step 1: log 3x+1 = log15 Step 2: (x + 1)log 3 = log15 Step 3: log3 = log 15 over x plus 1 Step 4: 0.477121 = 1.176091 over x plus 1 Step 5: 0.477121(x + 1) = 1.176091 Step 6: x + 1 = 1.176091 over 0.477121 Step 7: x + 1 = 2.464975 Step 8: x = 1.464975

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  1. anonymous
    • one year ago
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    @Hero

  2. Hero
    • one year ago
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    @alexistheking777 what are your thoughts regarding this problem? Do you see which step contains the error?

  3. anonymous
    • one year ago
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    I think it is either step 2 or 3?

  4. Hero
    • one year ago
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    By the way, if you use \(\LaTeX\) you'll be able to properly post the exponential equation. For example, you can create this: \(3^{x + 1} = 15\)

  5. anonymous
    • one year ago
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    Oh okay sorry but I think it is either 2 3 or 4 2 is my first choice 4 is my second 3 is my third

  6. Hero
    • one year ago
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    Let's go one step at a time. We know the first step is correct because we're allowed to log both sides. What do you think is possibly wrong with the second step?

  7. anonymous
    • one year ago
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    The fact that x+1 is multiplied to log 3

  8. Hero
    • one year ago
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    In order for any step to be wrong, we must be able to identify some mathematical rule that has been violated. Since we're using logs, then we should be able to say which rule of logs was violated for the second step. Do you know what that is (if any)?

  9. anonymous
    • one year ago
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    No, I'm sorry I have a hard time with logs.

  10. Hero
    • one year ago
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    How familiar does the following rule look to you? \(\log(a^b) = b \log(a)\)

  11. anonymous
    • one year ago
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    pretty familiar

  12. Hero
    • one year ago
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    Okay, now suppose \(a = 3\) and \(b = x+1\). How would using that rule of logs I posted above compare with what was done in step 2?

  13. anonymous
    • one year ago
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    the x+1 can't be multiplied in step 2 because it has to be an exponent i think

  14. anonymous
    • one year ago
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    @Hero

  15. Hero
    • one year ago
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    So for the rule \(\log(a^b) = b \log(a)\) How would you describe the role of \(b\) on the RIGHT hand side of the equation? Is \(b\) an exponent?

  16. anonymous
    • one year ago
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    no not on the right side but on the left side ya

  17. Hero
    • one year ago
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    \(b\) is not an exponent on the right side. Correct. So in other words, according to the rule, is it possible to re-write \(\log(a^b)\) so that \(b\) is not an exponent?

  18. anonymous
    • one year ago
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    Yes because it isn't on the right side

  19. Hero
    • one year ago
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    Okay, so knowing this, what can we say about step 2?

  20. anonymous
    • one year ago
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    We can say step 2 is incorrect

  21. anonymous
    • one year ago
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    Give me like 30 minutes @Hero I'll be back tho.

  22. Hero
    • one year ago
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    So for the rule\( log(a^b)=b \log(a)\) : b is allowed to be written as a multiple of log(a) but x+1 isn't allowed to be written as a multiple of log(3) ?

  23. Hero
    • one year ago
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    While you're taking a break, you should refer back to the part where I suggested for you to let \(a = 3\) and \(b = x + 1\) for the log rule I already posted.

  24. anonymous
    • one year ago
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    Hey @Hero I think it can be I think x + 1 can be written as a multiple of log(3)

  25. Hero
    • one year ago
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    Yes, I just showed it in the drawing above.

  26. Hero
    • one year ago
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    You might have to zoom out to see it all.

  27. anonymous
    • one year ago
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    I can't see the drawing for some reason. But with that being said it can't be one or two so I think step 4 would be incorrect. I think step 3 is right because she needed to get rid of x+1.

  28. Hero
    • one year ago
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    What do you mean "get rid of" x + 1? Why does she "need" to do that?

  29. anonymous
    • one year ago
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    Because I think you have to do that to simplify but maybe I'm thinking of easier algebra. :(

  30. anonymous
    • one year ago
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    @Hero

  31. Hero
    • one year ago
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    She "needs" to isolate "x". Dividing both sides by x + 1 is not a step in the direction of isolating x. Dividing both sides by x + 1 isn't getting rid of it. Instead, what it does is limit the domain of x which isn't the goal at all. I guess now you know which step is incorrect.

  32. anonymous
    • one year ago
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    Thank you can you help me with a couple more if that's alright with you? Your a genius

  33. anonymous
    • one year ago
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    @Hero

  34. Hero
    • one year ago
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    Post your next question as a new question (separate from this one). Remember anyone can help with a question.

  35. anonymous
    • one year ago
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    I tagged you in the other one @Hero

  36. Hero
    • one year ago
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    |dw:1435362088247:dw|

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