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7 January, 11:56

If (6^2) ^x = 1 what is the value of x

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  1. 7 January, 13:18
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    x = 0

    Step-by-step explanation:

    Rewrite (6^2) ^x = 1 as 6^ (2x) = 1.

    Next, take the logarithm (common or natural) of both sides, obtaining

    2x log 6 = log 1

    2x log 6 = 0

    Then x = 0.

    Note that (6^2) ^0 = 1, since any positive number raised to the prower 0 is 1.
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