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15 July, 17:19

How do you solve this equation 3^3x-7=81^12-3x

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  1. 15 July, 17:37
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    I think you meant ...

    3^ (3x-7) = 81^ (12-3x) (note that 81=3^4) so

    3^ (3x-7) = 3^4^ (12-3x)

    3^ (3x-7) = 3^ (48-12x) taking log or natural log of both sides show that

    3x-7=48-12x

    15x=55

    x=11/3
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