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Spring98

  • one year ago

Part A: If (2^6)^x = 1, what is the value of x? Explain your answer. Part B: If (5^0)^x = 1, what are the possible values of x? Explain your answer.

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  1. Spring98
    • one year ago
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    @jdoe0001 Can you help me with this one too. I can't pas my quiz and if i don't get 80% or more i can't go on.

  2. jdoe0001
    • one year ago
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    \(\bf \textit{part A, multiply both sides by }\qquad \cfrac{1}{26} \\ \quad \\ \textit{part B, multiply both sides by }\qquad \cfrac{1}{50}\) what does that give you?

  3. Spring98
    • one year ago
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    they both equal 1 @jdoe0001

  4. Spring98
    • one year ago
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    is that right?

  5. jdoe0001
    • one year ago
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    \(\bf (26)x=1\implies \cfrac{1}{\cancel{26}}\cdot (\cancel{26})x=1\cdot \cfrac{1}{26}\implies x=? \\ \quad \\ \quad \\ (50)x=1\implies \cfrac{1}{\cancel{50}}\cdot (\cancel{50})x=1\cdot \cfrac{1}{50}\implies x=?\)

  6. Spring98
    • one year ago
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    0.0384615384615385 is part A and 0.02for part B @jdoe0001

  7. jdoe0001
    • one year ago
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    \(\huge \circ.\circ\)

  8. jdoe0001
    • one year ago
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    notice, the number on the left cancelled out with the denominator

  9. Spring98
    • one year ago
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    but what happened with the exponents?

  10. Spring98
    • one year ago
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    @jdoe0001

  11. Spring98
    • one year ago
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    @zepdrix @jdoe0001 can you guys plz help me?

  12. jdoe0001
    • one year ago
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    woops, got a typo there \(\bf (26)x=1\implies \cfrac{1}{\cancel{26}}\cdot (\cancel{26})x=1\cdot \cfrac{1}{26}\implies \cfrac{1}{1}\cdot x=\cfrac{1}{26}\implies x=\cfrac{1}{26} \\ \quad \\ \quad \\ (50)x=1\implies \cfrac{1}{\cancel{50}}\cdot (\cancel{50})x=1\cdot \cfrac{1}{50}\implies \cfrac{1}{1}\cdot x=\cfrac{1}{50}\implies x=\cfrac{1}{50}\)

  13. Spring98
    • one year ago
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    i know but there is expopnents and i'm going to get this wrong.

  14. jdoe0001
    • one year ago
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    ohh...shoot.. is not what you originally posted though hmm.... need to dash in secs..... you may want to repost

  15. zepdrix
    • one year ago
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    \[\large\rm (2^{6})^x=1\]Applying a rule of exponents to the left side gives us\[\large\rm 2^{6x}=1\]Rewrite the right side as a power of 2. 2 to what power = 1?

  16. Spring98
    • one year ago
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    ok i will repost and i will tag you k @jdoe0001

  17. Spring98
    • one year ago
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    Is it 0? @zepdrix

  18. zepdrix
    • one year ago
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    Good. Anything to the zero power is 1. 2^0=1. Therefore we can write our equation:\[\large\rm 2^{6x}=1\]as\[\large\rm 2^{6x}=2^0\]Since the bases are the same, we can equate the exponents.

  19. Spring98
    • one year ago
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    how did the 2 end up a 0. and isn't the x supposed to be a 0

  20. Spring98
    • one year ago
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    @zepdrix

  21. zepdrix
    • one year ago
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    I didn't change anything about the left side. I rewrite 1 as 2^0.

  22. zepdrix
    • one year ago
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    Therefore, equating the exponents gives us \(\large\rm 6x=0\) which leads to \(\large\rm x=0\)

  23. zepdrix
    • one year ago
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    For the other problem:\[\large\rm (5^0)^x = \color{orangered}{1}\]again, rewrite 1 as a power of 5,\[\large\rm (5^0)^x = \color{orangered}{5^0}\]Apply exponent rule to the left side,\[\large\rm (5^{0x}) = \color{orangered}{5^0}\]Which leads to\[\large\rm 0x=0\]And it looks like ANY value of x is going to solve this one.

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