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Math Help - Simple indices question..

  1. #1
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    Simple indices question..

    Solve the equation 4^2x x 8^(x-1) = 32

    Usually I solve these using logs and antilogs but here the question specifies that I cannot use that way as it's too basic.

    I get the answer, x = 8/7 using logs/antilogs but I am not sure how to do it using a just the indices and surds rules.
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  2. #2
    Like a stone-audioslave ADARSH's Avatar
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    Quote Originally Posted by struck View Post
    Solve the equation 4^2x x 8^(x-1) = 32

    Usually I solve these using logs and antilogs but here the question specifies that I cannot use that way as it's too basic.

    I get the answer, x = 8/7 using logs/antilogs but I am not sure how to do it using a just the indices and surds rules.
    4^2x x 8^(x-1) = 32

    I think you mean
    4^{2x} \times 8^{x-1} = 32

    (2^2)^{2x} \times (2^3)^{x-1} = 2^5

    2^{4x} \times 2^{3x-3}= 2^5

    2^{4x+3x-3} = 2^5

    <br />
2^{7x-3} = 2^ 5

    Thus
    7x-3 = 5

    x= 8/7
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  3. #3
    MHF Contributor
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    Quote Originally Posted by struck View Post
    Solve the equation 4^2x x 8^(x-1) = 32

    Usually I solve these using logs and antilogs but here the question specifies that I cannot use that way as it's too basic.

    I get the answer, x = 8/7 using logs/antilogs but I am not sure how to do it using a just the indices and surds rules.
    4^{2x}\cdot 8^{x-1} = 32

     <br />
2^{4x}\cdot2^{3x-3}=2^5<br />
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