college algebra

solve the following logarithmic equation.
1. 2log(lower2) (x-2) + log(lower2)^2=3
I do not know how to make the 2 by the g on my computer.

2. log(lower729^x) + log(lower27^x) + log(lower3^x) =9
again I do not know how to make the numbers lower case

please show work, and if you can tell me how to make the lower case numbers on here.

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  1. lots of folks just write log_2(x-2)
    the numbers are subscripts, not lower-case.

    1. since we're dealing with logs base 2, just say log:

    2log(x-2) + log2 = 3
    log(x-2)^2 + log2 = 3
    log(2(x-2)^2) = 3
    2(x-2)^2 = 8
    (x-2)^2 = 4
    x = 0 or 4

    I assumed that log(lower2)^2 meant log(lower2)2

    2.
    since 729 = 3^5, log base 729 = 1/5 log base 3
    since 27 = 3^3, log base 27 = 1/3 log base 3
    so, using all logs base 3,

    1/6 logx + 1/3 logx + logx = 9
    3/2 logx = 9
    logx = 6
    x = 3^6 = 729

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