# 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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