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smurfy14

  • 5 years ago

I need help on how to solve this problem and how to set it up. 3^(3x-1)=9(x-3)

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  1. smurfy14
    • 5 years ago
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    Typo on last post it should be: 3^(3x-1)=9^(x-3)

  2. anonymous
    • 5 years ago
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    first u note that 9^(x-3) = 3^[2(x-3)] therefore; 3x-1=2(x-3). solve this to get x.

  3. smurfy14
    • 5 years ago
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    where did the two come from?

  4. anonymous
    • 5 years ago
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    thats because 3^2=9

  5. anonymous
    • 5 years ago
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    as example, 9^2=3^(2*2)=81. thats the same i done above. but instead of integer, the power in the question is an equation.

  6. smurfy14
    • 5 years ago
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    ok so if i have a problem like 3^(4x-2)=27^(x) then it would be 4x-2=3(x)??

  7. anonymous
    • 5 years ago
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    yup. u got it right.

  8. smurfy14
    • 5 years ago
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    now what if i got a^(x-4)=[a^6][a^(3x-6)]

  9. anonymous
    • 5 years ago
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    let say u hav (2^3)*(2^5), this wil be equal to 2^(3+5) so in ur case, [a^6][a^(3x-6)]=a^[6+(3x-6)]

  10. smurfy14
    • 5 years ago
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    so then you would do a^x-4=a^[6+(3x-6)] and then what do you do next?

  11. anonymous
    • 5 years ago
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    yup.then do same as before, equate the power to get the x. x-4=6+(3x-6)

  12. smurfy14
    • 5 years ago
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    alright now if you have a problem 9^(2x+1)=27^(x+2) then you have 2x+1=???

  13. anonymous
    • 5 years ago
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    in this case, first u need to make the "base" equal. note that 9=3^2 and 27=3^3. therefore,the eqn become: 3^[2(2x+1)]=3^[3(x+2)] then equate the power as usual: 2(2x+1)=3(x+2)

  14. smurfy14
    • 5 years ago
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    ok so now i have 5^(2x+3)=(square root of 5)^(x+4)

  15. anonymous
    • 5 years ago
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    u can write square root of 5 as 5^(1/2) then u can proceed with 2x+3= (1/2)*(x+4)

  16. smurfy14
    • 5 years ago
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    why do you write it as 1/2? how did you get that

  17. anonymous
    • 5 years ago
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    dont have to calculate anything for that. its basic rule how to write it a square root. \[\sqrt[m]{A} = A ^{1/m}\]

  18. smurfy14
    • 5 years ago
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    ohh ok haha also, i have (2a)^x=(4a^2)^x+2 and so would it be x=2(x+2)?

  19. anonymous
    • 5 years ago
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    yes.thats right.

  20. anonymous
    • 5 years ago
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    if the ratio of snicker to m&m's is 5 to 6 then the store could have

  21. anonymous
    • 5 years ago
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    help

  22. smurfy14
    • 5 years ago
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    and on the last problem with the 1/2... what would the answer be?

  23. anonymous
    • 5 years ago
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    i think it is x=-2/3. u got the same?

  24. anonymous
    • 5 years ago
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    can u help me on my pro

  25. smurfy14
    • 5 years ago
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    no i got something different :/ umm i guess i did it wrong because i think you are right

  26. anonymous
    • 5 years ago
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    try check it again. maybe a careless mistake.

  27. smurfy14
    • 5 years ago
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    ok now i got it :) okay well to switch gears a bit.. now i have to do log(3/5)(25/9)=x+4 **3/5 is the little bottom number**

  28. anonymous
    • 5 years ago
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    remember that if \[\log_{a} b = c \] this implies\[a ^{c}=b\] so, u hav to change the initial problem into the 2nd form. and u also hav to note that 25/9= (3/5)^-2

  29. smurfy14
    • 5 years ago
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    you mean 5/3^-2?

  30. anonymous
    • 5 years ago
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    no. 25/9= (5/3)^2 = (3/5)^-2

  31. smurfy14
    • 5 years ago
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    ohh ok! im sorry to be asking so many questions its just that im so confused! ok on 2^x * 4^(x+5)=4^(2x-1) what do you do next?

  32. anonymous
    • 5 years ago
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    since 4=2^2, then the eqn becomes; 2^x * 2^[2(x+5)]=2^[2(2x-1)] this implies; 2^[x+2(x+5)]=2^[2(2x-1)] therefore, x+2(x+5)=2(2x-1)

  33. smurfy14
    • 5 years ago
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    ohh ok umm what did you get on the previous problem? i dont know where x is supposed to go when you solve. i got this: (3/5)^4=(3/5)^-2

  34. anonymous
    • 5 years ago
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    it should be (3/5)^(x+4)=(3/5)^-2 x included in "c" based on the general form i gave u before.

  35. anonymous
    • 5 years ago
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    im going to bed.sorry cant help more.good luck with the problems..

  36. smurfy14
    • 5 years ago
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    ohh i forgot the x! so it should be -6.

  37. smurfy14
    • 5 years ago
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    ok thanks for all your help!!!! :)

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