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Adam1994

  • 4 years ago

Confused?? O.o

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  1. Adam1994
    • 4 years ago
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    |dw:1332877354175:dw|

  2. Adam1994
    • 4 years ago
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    Someone help meee

  3. jagatuba
    • 4 years ago
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    \[\left( \sqrt[4]{6} \right)^{12x-4}=36^{x+4}\] Is this correct?

  4. Adam1994
    • 4 years ago
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    yes thats it

  5. jagatuba
    • 4 years ago
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    I'll help you shortly.

  6. Adam1994
    • 4 years ago
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    okay thanks

  7. phi
    • 4 years ago
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    you do know \[\sqrt[4]{6}= 6^{\frac{1}{4}}\]

  8. phi
    • 4 years ago
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    you also know the general rule \[ (a^b)^c = a^{bc} \]

  9. phi
    • 4 years ago
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    so let's use both ideas. replace fourth root of 6 with 6^(1/4) \[ (6^{\frac{1}{4}})^{12x-4} = 6^{\frac{12x-4}{4} }\]

  10. phi
    • 4 years ago
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    can you simplify from here?

  11. Adam1994
    • 4 years ago
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    Yeah I think so, and it comes out to be x=9 correct?

  12. phi
    • 4 years ago
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    yes if we equate \[ 6^{3x-1} = 6^{2(x+4)} \] then the exponents must be equal 3x-1= 2x+8 x= 9

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