9514 1404 393
Answer:
g(x) = 1/2
Step-by-step explanation:
The applicable relations are ...
[tex]\log_b(x)=a\ \Longleftrightarrow\ b^a=x\\\\\log_b(x^a)=a\cdot\log_b(x)\\\\\log_b(b)=1[/tex]
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We want to find logb(x) for x = b^(1/2). Using the first relation, we see ...
g(x) = logb(x)
b^(g(x)) = b^(1/2) . . . . . using g(x) = a
g(x) = 1/2 . . . . . equating exponents
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Alternatively, using the last two relations, we have x = b^(1/2), so ...
g(b^(1/2)) = logb(b^(1/2)) = (1/2)logb(b) = (1/2)(1)
g(x) = 1/2
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Additional comment
I find it useful to remember that a logarithm is an exponent. The first applicable relation shown above is a statement to that effect.
