DarkBlueChocobo
  • DarkBlueChocobo
Help with logarithims
Mathematics
katieb
  • katieb
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DarkBlueChocobo
  • DarkBlueChocobo
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DarkBlueChocobo
  • DarkBlueChocobo
For the first one would you just fill in x for 0?
DarkBlueChocobo
  • DarkBlueChocobo
to which we would get log(10)10+3

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freckles
  • freckles
yes replace x with 0
freckles
  • freckles
\[\log_a(a)=1 \text{ where } a \in (0,1) \cup (1,\infty)\]
freckles
  • freckles
since a^1 is a
DarkBlueChocobo
  • DarkBlueChocobo
I am confused are you saying log10(10)=1?
freckles
  • freckles
yes \[10 \in (1,\infty) \\ \text{ so } 10 \in (0,1) \cup (1,\infty) \\ \text{ so yes } \log_{10}(10)=1 \]
DarkBlueChocobo
  • DarkBlueChocobo
so like then we just add 3? so f(0)=4?
freckles
  • freckles
since 10^1=10
freckles
  • freckles
yep
DarkBlueChocobo
  • DarkBlueChocobo
and for the second one would you just divide log(10)20/log(10)3
freckles
  • freckles
yes
DarkBlueChocobo
  • DarkBlueChocobo
so log(3)20=x so you just seperate? and then they both have the common log(10)? im trying to udnerstand why
freckles
  • freckles
you can solve the equation by first taking log_10 of both sides
freckles
  • freckles
\[\log_{10}(20)=\log_{10}(3^x) \\ \log_{10}(20)=x \log_{10}(3) \text{ by power rule } \]
freckles
  • freckles
then divide both sides be log_10(3)
DarkBlueChocobo
  • DarkBlueChocobo
Alrights Thank you for clearing that up. Egh last night of cramming for this test
freckles
  • freckles
\[\frac{\log_{10}(20)}{\log_{10}(3)}=\frac{x \log_{10}(3)}{\log_{10}(3)} \\ \frac{\log_{10}(20)}{\log_{10}(3)}=x\]
freckles
  • freckles
you could have chosen any log base as long as that base is a number between 0 and 1 exclusive or between 1 and infinity exclusive
freckles
  • freckles
maybe they want you to express in base 5 log well take log_5( ) of both side instead
freckles
  • freckles
just doing an example... \[20=3^x \\ \log_5(20)=\log_5(3^x) \\ \log_5(20)=x \log_5(3) \\ \frac{\log_5(20)}{\log_5(3)}=x\]
freckles
  • freckles
notice that answer could be expressed as: \[x=\frac{\log_a(20)}{\log_a(3)} \text{ where } a \in (0,1) \cup (1,\infty)\]
anonymous
  • anonymous
can we say 3^x=20?
freckles
  • freckles
yes equality is symmetrical if a=b then b=a
anonymous
  • anonymous
the can we say x=log3(20)?
freckles
  • freckles
yes my answer will also conclude that if you put a as 3 \[x=\frac{\log_3(20)}{\log_3(3)}=\frac{\log_3(20)}{1}=\log_3(20)\]
anonymous
  • anonymous
can't we say x as a decimal?
freckles
  • freckles
you mean approximate x ?
anonymous
  • anonymous
yep
anonymous
  • anonymous
3^1=3 3^2=9 3^3=27 so x must be between 2 and three.... and so on?
anonymous
  • anonymous
But if we are asked to find x for 4 decimal places can we find it? :(
freckles
  • freckles
calculator would probably be your friend
anonymous
  • anonymous
can we find it without using it?

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