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AS Level Maths Core 1 indices question- Help! Watch

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    hi everyone, was wondering if anyone could help me out with this maths question thats been bothering me, any advice how to tackle it would be appreciated greatly
    thanks!
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    Here's the question


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    Here's the solution. Do loads of practice. Indices will get you, practice will over come that!

    Just make sure the you're using the same base, I saw that I had the opportunity to use base 3 as it can be adapted to both expressions. Since both sides have the same base, the sum of each sides indices will equal each other, as shown.

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    (Original post by TheTechnoGuy)
    Here's the solution. Do loads of practice. Indices will get you, practice will over come that!

    Just make sure the you're using the same base, I saw that I had the opportunity to use base 3 as it can be adapted to both expressions. Since both sides have the same base, the sum of each sides indices will equal each other, as shown.

    Name:  Picture2.jpg
Views: 119
Size:  239.4 KB
    Thank you!! Very helpful


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    TheTechnoGuy, the method you've followed is correct but you have misread the question. You should end up with the equation 2x = 3-3x, leading to x=3/5.
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    Darn.... my bad. Been working with numbers all day so its bound to happen :P
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    Happens to the best of us
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    (Original post by TheTechnoGuy)
    Here's the solution. Do loads of practice. Indices will get you, practice will over come that!

    Just make sure the you're using the same base, I saw that I had the opportunity to use base 3 as it can be adapted to both expressions. Since both sides have the same base, the sum of each sides indices will equal each other, as shown.

    Name:  Picture2.jpg
Views: 119
Size:  239.4 KB
    full solutions are against forum guidelines


    (Original post by holly mcnally)
    hi everyone, was wondering if anyone could help me out with this maths question thats been bothering me, any advice how to tackle it would be appreciated greatly
    thanks!
    You have

    \displaystyle 9^{x} = 27^{1-x}

    This can be rewritten as

    \displaystyle \left(3^{2}\right)^{x} = \left(3^{3}\right)^{1-x}\]

    Can you work from here?
 
 
 
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