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    How would i solve,

    4x-3(2x+1)=0
    thanks =)

    I have no working as I don't even know where to start, =(
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    HINT:

    -  \log(a) + \log(b) = \log(ab)
    -  \log(a) - \log(b)  = \log \left(\frac{a}{b} \right)
    - \log(a^b) = b \log (a)
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    (Original post by Hi, How are you ?)
    How would i solve,

    4x-3(2x-1)=0
    thanks =)

    I have no working as I don't even know where to start, =(
    This reduces to:

    \displaystyle 2^{2x} - \frac{3}{2} 2^x = 0

    This is a quadratic equation that can be factorised.

    Note that 2^x =0 does not have a solution. Why?
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    (Original post by Mr M)
    This reduces to:

    \displaystyle 2^{2x} - \frac{3}{2} 2^x = 0

    This is a quadratic equation that can be factorised.

    Note that 2^x =0 does not have a solution. Why?
    sorry, it was supoposed to be

    4x-3(2x+1), not 4x-3(2x-1)
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    (Original post by Hi, How are you ?)
    sorry, it was supoposed to be

    4x-3(2x+1), not 4x-3(2x-1)
    Ok then the equation is:

    2^{2x}-6(2^x)=0
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    (Original post by Mr M)
    Ok then the equation is:

    2^{2x}-6(2^x)=0
    so,

    4x = (22)x =22x according the the laws of indicies.

    then 3(2x+1) = 3(2x ^ 21) = 6(2x) (i don't understand the last part, what happens to the 21 ?
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    (Original post by Hi, How are you ?)
    so,

    4x = (22)x =22x according the the laws of indicies.

    then 3(2x+1) = 3(2x ^ 21) = 6(2x) (i don't understand the last part, what happens to the 21 ?
    You are mistaken.

    2^{x+1}=2^x \times 2^1
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    (Original post by Mr M)
    You are mistaken.

    2^{x+1}=2^x \times 2^1
    so, in fact you are only multiplying the powers, not the 2, then ?
    but why would you do this?
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    (Original post by Hi, How are you ?)
    so, in fact you are only multiplying the powers, not the 2, then ?
    but why would you do this?
    I don't know what you are asking? Can you reword your question?
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    (Original post by Mr M)
    I don't know what you are asking? Can you reword your question?
    Your stating that 2x+1 = (2x times 21) = 2x, right?

    I don't understand how (2x times 21) would euqal 2x
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    (Original post by Hi, How are you ?)
    Your stating that 2x+1 = (2x times 21) = 2x, right?

    I don't understand how (2x times 21) would euqal 2x
    I am not stating that.

    I am saying that 3 \times 2^{x+1} = 3 \times 2^1 \times 2^x = 6 \times 2^x
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    (Original post by Mr M)
    I am not stating that.

    I am saying that 3 \times 2^{x+1} = 3 \times 2^1 \times 2^x = 6 \times 2^x
    OK, what happens to the "21" part, should is not times the whole lot by 2?
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    (Original post by Hi, How are you ?)
    OK, what happens to the "21" part, should is not times the whole lot by 2?
    2^1 = 2

    so

     3 \times 2^1 \times2^x = 3 \times 2 \times 2^x = 6 \times 2^x .
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    OK, i see, thanks for the help, +ve reps
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    (Original post by Hi, How are you ?)
    OK, what happens to the "21" part, should is not times the whole lot by 2?
    3 x 2 = 6 ?

    I see you understand that bit now. Have you solved the original question?
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    (Original post by Mr M)
    3 x 2 = 6 ?

    I see you understand that bit now. Have you solved the original question?
    (22)x-3(2 times 2x)

    then, 22x - 6(2x)=0

    I let y = 2x

    and got y2-6y=0
    factorized and got y=0 and y=6, but y can't be 0 as not soloutions to 0=2x

    the i got 6=2x, then x= Log(6) / Log(2) = 2.58 (3sf)
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    (Original post by Hi, How are you ?)
    (22)x-3(2 times 2x)

    then, 22x - 6(2x)=0

    I let y = 2x

    and got y2-6y=0
    factorized and got y=0 and y=6, but y can't be 0 as not soloutions to 0=2x

    the i got 6=2x, then x= Log(6) / Log(2) = 2.58 (3sf)
    Yes - the slightly easier way to solve 2^x =6 is to realise that x=\log_2 6 .
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    (Original post by Mr M)
    Yes - the slightly easier way to solve 2^x =6 is to realise that x=\log_2 6 .
    I see, thanks. One more thing, what do people mean when they say, "taking logs of both sides". Can you please explain and possible provide me with an example to see it in action?
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    (Original post by Hi, How are you ?)
    I see, thanks. One more thing, what do people mean when they say, "taking logs of both sides". Can you please explain and possible provide me with an example to see it in action?
    Yes I think you already know how to do this.

    To solve 2^x = 6 by taking logs of both sides.

    \log 2^x = \log 6

    x \log 2 = \log 6

    \displaystyle x=\frac{\log 6}{\log 2}

    That's it!
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    (Original post by Mr M)
    Yes I think you already know how to do this.

    To solve 2^x = 6 by taking logs of both sides.

    \log 2^x = \log 6

    x \log 2 = \log 6

    \displaystyle x=\frac{\log 6}{\log 2}

    That's it!
    So it is just a case of adding log to each item in the equation then, to the base 10?
 
 
 
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