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penelopecrux
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#1
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#1
idk how you get y=10^3log+2 according to the mark scheme?
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_gcx
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(Original post by penelopecrux)
idk how you get y=10^3log+2 according to the mark scheme?
Can you post your attempt?
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RDKGames
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(Original post by penelopecrux)
idk how you get y=10^3log+2 according to the mark scheme?
The straight line would be an equation in the form \log(y)-b=m[\log(x)-a] where m is the gradient and (a,b) a point that it passes through. Then just solve for y.
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username2769500
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Simply you have the equation of a straight line y = mx + c then you replace x,m and c with the equal terms from the graph. Leave y because the questions answer has y and not log10y as the function.
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penelopecrux
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#5
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(Original post by _gcx)
Can you post your attempt?
this is my attempt
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_gcx
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(Original post by penelopecrux)
this is my attempt
How can we "cancel" the logarithm on the LHS, to write the equation in the form y=f(x)?
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penelopecrux
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(Original post by _gcx)
How can we "cancel" the logarithm on the LHS, to write the equation in the form y=f(x)?
would u write the 2 as log100?
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RDKGames
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(Original post by penelopecrux)
would u write the 2 as log100?
You could, but no point. Just think of an operation that gets rid off the log on the LHS, and apply it on both sides.
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penelopecrux
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(Original post by RDKGames)
You could, but no point. Just think of an operation that gets rid off the log on the LHS, and apply it on both sides.
i'm not rlly sure tbh, all i can think to do is just to put the 3 up as a power to the x
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RDKGames
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(Original post by penelopecrux)
i'm not rlly sure tbh, all i can think to do is just to put the 3 up as a power to the x
What's the reverse operation of taking the logarithm??

If I wanted to find x in 2^x=8 and I would take the logarithm of both sides which gives \log(2^x)=\log(8)

So the reverse operation is to...?
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penelopecrux
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(Original post by RDKGames)
What's the reverse operation of taking the logarithm??

If I wanted to find x in 2^x=8 and I would take the logarithm of both sides which gives \log(2^x)=\log(8)

So the reverse operation is to...?
take logs of both sides?
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RDKGames
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(Original post by penelopecrux)
take logs of both sides?
No, you'd exponentiate both sides...

If I go back from \log(2^x) I don't take logs to both sides, because then I'd have \log(\log(2^x)).

Instead you need to take the expression and put it as the exponent to the base of the logarithm. So \log(2^x)=\log(8) \Rightarrow 10^{\log(2^x)}=10^{\log(8)} and it simplifies from there. (note that \log is the same as \log_{10})

You may wish to review this in your chapter.
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penelopecrux
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(Original post by RDKGames)
No, you'd exponentiate both sides...

If I go back from \log(2^x) I don't take logs to both sides, because then I'd have \log(\log(2^x)).

Instead you need to take the expression and put it as the exponent to the base of the logarithm. So \log(2^x)=\log(8) \Rightarrow 10^{\log(2^x)}=10^{\log(8)} and it simplifies from there. (note that \log is the same as \log_{10})

You may wish to review this in your chapter.
yeah but how do i do that with the equation i have currently?
logy=logx^3 +2)
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_gcx
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(Original post by penelopecrux)
yeah but how do i do that with the equation i have currently?
logy=logx^3 +2)
If we exponentiate x (with base 10), then we have 10^x. How can we therefore exponentiate both sides of this equation?
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penelopecrux
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#15
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(Original post by _gcx)
If we exponentiate x (with base 10), then we have 10^x. How can we therefore exponentiate both sides of this equation?
i rly dont know
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RDKGames
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(Original post by penelopecrux)
i rly dont know
10^{\log_{10}(y)}=y=10^{3\log_{1  0}(x)+2} and simplify RHS.

Again, you may wish to review this chapter - this is a very basic principle.
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penelopecrux
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(Original post by RDKGames)
10^{\log_{10}(y)}=y=10^{3\log_{1  0}(x)+2} and simplify RHS.

Again, you may wish to review this chapter - this is a very basic principle.
yeah i'll definitely give it a review

i just don't get how logy turns into that??
apologies for all this
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RDKGames
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(Original post by penelopecrux)
yeah i'll definitely give it a review

i just don't get how logy turns into that??
apologies for all this
\displaystyle a^b=c \Rightarrow \log_a(a^b)=\log_a(c) \Rightarrow b \underbrace{\log_a(a)}_{=1}=\log  _a(c)

so b=\log_a(c)

Happy with this??

Now start with the RHS

b=\log_a(c) \Rightarrow a^b=a^{\log_a(c)} \Rightarrow a^b=c=a^{\log_a(c)} which in other words means a^{\log_a(c)}=c
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