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

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nowheremanBest ResponseYou've already chosen the best response.0
So how far do you get?
 2 years ago

animalsavior94Best ResponseYou've already chosen the best response.0
not at all because its so confusing
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
is this log base 5?
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
actually it doesnt matter so lets just write log
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
and ignore the 1/3 out front we will deal with it last
 2 years ago

animalsavior94Best ResponseYou've already chosen the best response.0
yes its a base 5
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
first use \[\log(a^n)=n\log(a)\] backwards. all the multipliers come inside the log as exponents
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
\[4\log(x+1)=\log((x+1)^4)\] \[3\log(3x)=\log((3x)^3)\] \[6\log(x)=\log(x^6)\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
giving \[\log((x+1)^4)\log((3x)^3)+\log(x^6)\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
now we use \[\log(a)log(b)=log(\frac{a}{b})\]
 2 years ago

animalsavior94Best ResponseYou've already chosen the best response.0
what happened to the 1/3 outside the []
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
i will deal with that last
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
\[log((x+1)^4)\log((3x)^3)=\log(\frac{(x+1)^4}{(3x)^3})\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
now we use \[\log(a)+\log(b)=\log(ab)\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
\[\log(\frac{(x+1)^4}{(3x)^3})+\log(x^6)=\log(\frac{x^6(x+1)^4}{(3x)^3})\]
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
now for the 1/3 out front. take the cube root of all that stuff.
 2 years ago

satellite73Best ResponseYou've already chosen the best response.1
the inner stuff i mean. the cube root of x^6 is x^2, and the cube root of (3x)^3 is 3x and so you get \[\log(\frac{x^2(x+1)^{\frac{4}{3}}}{3x})\]
 2 years ago

animalsavior94Best ResponseYou've already chosen the best response.0
wow that was a hard problem but thanks again:)
 2 years ago
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