Question

using the Law of Logarithms to expand the expression..

Please help!

Answer 1

\[\log \left( \frac{ 10^{x} }{ x(x ^{2}+1)(x ^{4}+2) } \right)\]

Answer 2

have you considered first using the quotient rule for log?
\[\log(\frac{u}{v})=\log(u)-\log(v) \\ \]

Answer 3

\[\log(uv)=\log(u)+\log(v) \text{ is another rule you can use after that } \\ \log(u^r)=rlog(u) \text{ is the power rule which can also be used here }\]

Answer 4

\[\log10^{x}-logx (x ^{2}+1)+\log(x ^{4}+2)\]

Answer 5

I think you forgot to distribute the minus sign

Answer 6

\[xlog10-logx(x ^{2}+1)+\log(x ^{4}+2)\]

Answer 7

but since the denominator is looks like multiplication wouldn't be a addition?

Answer 8

\[\log(\frac{10^x}{x(x^2+1)(x^4+2)}) \\ \log(10^x)-\log(x(x^2+1)(x^4+2)) \\ \log(10^x)-[\log(x(x^2+1)(x^4+2)) ] \\ \log(10^x)-[\log(x(x^2+1))+\log(x^4+2)]] \\ \log(10^x)-\log(x(x^2+1))-\log(x^4+2) \]

Answer 9

though could also expand the log(x(x^2+1)) part too

Answer 10

ooh so the signs changed because of the minus sign before the brackets?

Answer 11

yep that is called the distributive property

Answer 12

oh ok.. sorry i suck on paying attention to small details.

Answer 13

ok so could i expand more after log(x(x^2+1))−log(x^4+2)?

Answer 14

the log(x(x^2+1)) could be expanded more

Answer 15

notice x(x^2+1) is a product of x and (x^2+1)

Answer 16

would i t be: Logx+log (x^2+1)

Answer 17

yes and don't forget we also have a - sign in front of the log(x(x^2+1)
so you have:
\[\log(\frac{10^x}{x(x^2+1)(x^4+2)}) \\ \log(10^x)-\log(x(x^2+1)(x^4+2)) \\ \log(10^x)-[\log(x(x^2+1)(x^4+2)) ] \\ \log(10^x)-[\log(x(x^2+1))+\log(x^4+2)]] \\ \log(10^x)-\log(x(x^2+1))-\log(x^4+2) \\ \log(10^x)-[\log(x)+\log(x^2+1)]-\log(x^4+2)\]

Answer 18

anyways yes log(10^x) =xlog(10)
but assuming this is log base 10 you could also simplify the x log(10) more

Answer 19

it will just become x right?

Answer 20

right

Answer 21

since log 10=1

Answer 22

yes

Answer 23

the result is: \[x-\log (x)-\log (x ^{2}+1)-\log (x ^{4}-2)\]

Answer 24

?

Answer 25

well you change the sign between x^4 and 2 for some reason

Answer 26

oh haha yeah.. oops

Answer 27

\[\log(\frac{10^x}{x(x^2+1)(x^4+2)}) \\ \log(10^x)-\log(x(x^2+1)(x^4+2)) \\ \log(10^x)-[\log(x(x^2+1)(x^4+2)) ] \\ \log(10^x)-[\log(x(x^2+1))+\log(x^4+2)]] \\ \log(10^x)-\log(x(x^2+1))-\log(x^4+2) \\ \log(10^x)-[\log(x)+\log(x^2+1)]-\log(x^4+2) \\ x \log(10)-\log(x)-\log(x^2+1)-\log(x^4+2) \\ x(1)-\log(x)-\log(x^2+1)-\log(x^4+2) \\ x-\log(x)-\log(x^2+1)-\log(x^4+2)\]
should be right if the log's are in base 10