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write as logarithm of a single quantity: 1/3[4log5(x+1)-3log5(3-x)+6log5x]

Mathematics
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So how far do you get?
not at all because its so confusing
is this log base 5?

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Other answers:

actually it doesnt matter so lets just write log
and ignore the 1/3 out front we will deal with it last
yes its a base 5
first use \[\log(a^n)=n\log(a)\] backwards. all the multipliers come inside the log as exponents
\[4\log(x+1)=\log((x+1)^4)\] \[3\log(3-x)=\log((3-x)^3)\] \[6\log(x)=\log(x^6)\]
giving \[\log((x+1)^4)-\log((3-x)^3)+\log(x^6)\]
now we use \[\log(a)-log(b)=log(\frac{a}{b})\]
what happened to the 1/3 outside the []
i will deal with that last
not done yet!
\[log((x+1)^4)-\log((3-x)^3)=\log(\frac{(x+1)^4}{(3-x)^3})\]
now we use \[\log(a)+\log(b)=\log(ab)\]
\[\log(\frac{(x+1)^4}{(3-x)^3})+\log(x^6)=\log(\frac{x^6(x+1)^4}{(3-x)^3})\]
now for the 1/3 out front. take the cube root of all that stuff.
the inner stuff i mean. the cube root of x^6 is x^2, and the cube root of (3-x)^3 is 3-x and so you get \[\log(\frac{x^2(x+1)^{\frac{4}{3}}}{3-x})\]
wow that was a hard problem but thanks again:)

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