Pre Calculus

Evaluate log b {square root of 10b}, given that log b 2 = 0.3562 and
log b 5 = 0.8271 (sorry I had no clue how to type this in math format!)

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  1. I am not sure if I interpret your question correctly.
    Given logb2 = 0.3562 and
    logb5 = 0.8271,
    evaluate logb{square-root of 10b};

    logb{square-root of 10b}
    = logb{square-root of 2b*5b}
    =(1/2)logb(10b)
    =(1/2)(1)
    =0.5
    since logb10b = b for any b.
    However, if the number 10 is replaced by 10 to the base 10, we proceed slightly differently:
    logb{square-root of 10}
    =(1/2)logb(10)
    = (1/2)logb(2*5)
    = (1/2)(logb(2)+logb(5))
    = (1/2)(0.3562+0.8271)
    =0.5916

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  2. I saw the question this way

    logb (√(10b))
    = 1/2(logb(10b))
    = 1/2(logb5 + logb2 + logbb)
    = 1/2(.8271 + .3562 + 1)
    = 1.0917

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  3. The second answer is the correct answer! Thanks to you both I am on my way to the final

    Shellie

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