3psilon
I am terrible with Logs plz help



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3psilon
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\[f(x) = \log_{3}(\sqrt(x) + 3 \ \]

3psilon
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Log base 3 square root x plus 3

3psilon
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How do you find the inverse of this/

anonymous
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start with the fact that \(\log_3(\sqrt{x})=\frac{1}{2}\log_3(x)\)

anonymous
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is it the log of the whole thing?

3psilon
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No just log base 3 square root x

anonymous
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i mean is it
\[f(x)=\log_3(\sqrt{x}+3)\] or
\[f(x)=\log_3(\sqrt{x})+3\]

3psilon
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2nd one

anonymous
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ok then start with
\[f(x)=\frac{1}{2}\log_3(x)+3\]

anonymous
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we can do the usual trick of writing \[y=\frac{1}{2}\log_3(x)+3\]then
\[x=\frac{1}{2}\log_3(y)+3\] and solve for \(y\)

anonymous
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you need the steps?

3psilon
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Yes please . I always get confused when there's stuff like log base 3

anonymous
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1) subtract 3
2) multiply by 2
3) raise 3 to everything

anonymous
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\[x=\frac{1}{2}\log_3(y)+3\]
\[x3=\frac{1}{2}\log_3(y)\]
\[2x6=\log_3(y)\]
\[3^{2x6}=y\]

anonymous
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last step because
\[\log_b(y)=x\iff b^x=y\]

3psilon
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Ohhh!!! Thank you so much !

anonymous
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yw