animalsavior94
write as logarithm of a single quantity: 1/3[4log5(x+1)3log5(3x)+6log5x]



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nowhereman
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So how far do you get?

animalsavior94
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not at all because its so confusing

anonymous
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is this log base 5?

anonymous
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actually it doesnt matter so lets just write log

anonymous
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and ignore the 1/3 out front we will deal with it last

animalsavior94
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yes its a base 5

anonymous
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first use
\[\log(a^n)=n\log(a)\] backwards. all the multipliers come inside the log as exponents

anonymous
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\[4\log(x+1)=\log((x+1)^4)\]
\[3\log(3x)=\log((3x)^3)\]
\[6\log(x)=\log(x^6)\]

anonymous
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giving
\[\log((x+1)^4)\log((3x)^3)+\log(x^6)\]

anonymous
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now we use \[\log(a)log(b)=log(\frac{a}{b})\]

animalsavior94
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what happened to the 1/3 outside the []

anonymous
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i will deal with that last

anonymous
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not done yet!

anonymous
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\[log((x+1)^4)\log((3x)^3)=\log(\frac{(x+1)^4}{(3x)^3})\]

anonymous
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now we use
\[\log(a)+\log(b)=\log(ab)\]

anonymous
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\[\log(\frac{(x+1)^4}{(3x)^3})+\log(x^6)=\log(\frac{x^6(x+1)^4}{(3x)^3})\]

anonymous
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now for the 1/3 out front. take the cube root of all that stuff.

anonymous
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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
\[\log(\frac{x^2(x+1)^{\frac{4}{3}}}{3x})\]

animalsavior94
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wow that was a hard problem but thanks again:)