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    I am really stuck on 5c and i have got the mark scheme but I dont understand it . any help would be appreciated
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    (Original post by D_Breezy)
    I am really stuck on 5c and i have got the mark scheme but I dont understand it . any help would be appreciated
    Which bit don't you undertstand?

    We can write \left(\frac{1}{3} \right)^x = \frac{1}{3^x}, so if we equate the two equations we have \frac{1}{3^x} = 2 3^x

    Multiply both sides by the denominator of the LHS:

    1 = 3^x \times 2 \times 3^x but you know what the same thing multiplied by itself is, don't you? It's a square. So:

    1 = 2 \times (3^x)^2. Now divide both sides by 2:

    \frac{1}{2} = (3^x)^2. Now take the square root:

    \pm \sqrt{\frac{1}{2}} = 3^x. reject the negative value because exponentials are always positive.

    SO 3^x = \sqrt{1/2}. Take logarithms to solve: x \ln 3 = \ln \sqrt{1/2}. Isolate x.
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    (Original post by Zacken)
    Which bit don't you undertstand?

    We can write \left(\frac{1}{3} \right)^x = \frac{1}{3^x}, so if we equate the two equations we have \frac{1}{3^x} = 2 3^x

    Multiply both sides by the denominator of the LHS:

    1 = 3^x \times 2 \times 3^x but you know what the same thing multiplied by itself is, don't you? It's a square. So:

    1 = 2 \times (3^x)^2. Now divide both sides by 2:

    \frac{1}{2} = (3^x)^2. Now take the square root:

    \pm \sqrt{\frac{1}{2}} = 3^x. reject the negative value because exponentials are always positive.

    SO 3^x = \sqrt{1/2}. Take logarithms to solve: x \ln 3 = \ln \sqrt{1/2}. Isolate x.
    hey thanks i just forgot
    . thank you !
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    (Original post by D_Breezy)
    hey thanks i just forgot
    . thank you !
    Yep, just remember that \left(\frac{a}{b}\right)^c = \frac{a^c}{b^c} and that in this case, a^c = 1^c = 1 since one to the power of anything is 1.
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    (1/3)^x =2(3)^x
    1/3=3^(-1)
    thus (3)^(-x)=2(3)^x
    multiplied by (3)^x on both sides
    then 1= 2(3)^(2x)
    (3)^(2x)=1/2
    2x=lg(1/2)/lg(3)
    x=1/2lg(1/2)/lg(3)
    using the calculator,x=-0.32
    then substitute x= 1/2lg(1/2)/lg(3)into y=(1/3)^x,1/2lg(1/2)/lg(3)=1/2log3(1/2)=log3(square root of 1/2)
    then y= (1/3)^x= 3^(-x)=3^(-log3(square root of 1/2)=3^log3(square root of 2)=root 2


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    what exam board do you do?
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    Looks like an Edexcel question.
 
 
 
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