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3psilon Group TitleBest ResponseYou've already chosen the best response.0
\[f(x) = \log_{3}(\sqrt(x) + 3 \ \]
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Log base 3 square root x plus 3
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
How do you find the inverse of this/
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
start with the fact that \(\log_3(\sqrt{x})=\frac{1}{2}\log_3(x)\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
is it the log of the whole thing?
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
No just log base 3 square root x
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i mean is it \[f(x)=\log_3(\sqrt{x}+3)\] or \[f(x)=\log_3(\sqrt{x})+3\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
ok then start with \[f(x)=\frac{1}{2}\log_3(x)+3\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
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 \(y\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you need the steps?
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Yes please . I always get confused when there's stuff like log base 3
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
1) subtract 3 2) multiply by 2 3) raise 3 to everything
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
\[x=\frac{1}{2}\log_3(y)+3\] \[x3=\frac{1}{2}\log_3(y)\] \[2x6=\log_3(y)\] \[3^{2x6}=y\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
last step because \[\log_b(y)=x\iff b^x=y\]
 2 years ago

3psilon Group TitleBest ResponseYou've already chosen the best response.0
Ohhh!!! Thank you so much !
 2 years ago
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