Question

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

Answer 1

So how far do you get?

Answer 2

not at all because its so confusing

Answer 3

is this log base 5?

Answer 4

actually it doesnt matter so lets just write log

Answer 5

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

Answer 6

yes its a base 5

Answer 7

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

Answer 8

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

\[3\log(3-x)=\log((3-x)^3)\]
\[6\log(x)=\log(x^6)\]

Answer 9

giving

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

Answer 10

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

Answer 11

what happened to the 1/3 outside the []

Answer 12

i will deal with that last

Answer 13

not done yet!

Answer 14

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

Answer 15

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

Answer 16

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

Answer 17

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

Answer 18

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

Answer 19

wow that was a hard problem but thanks again:)