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C2 - Logarithms

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Original post by RDKGames
Core 1 acts as the basis to Core 2, 3 and 4 so yes, you do.

Alternatively, you can complete the square to find the minimum point of that quadratic.


You can also find the maximum point using completing the square right?

Original post by RDKGames
Took them on my iphone, uploaded them to my computer. They are JPG format.


Just use tinypic or Imageshack lol.
Original post by Chittesh14
You can also find the maximum point using completing the square right?


Yes.
Original post by RDKGames
Yes.


How is this an example of the change of base rule in logarithms. I never understood this rule, but yeah...
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Original post by Chittesh14
How is this an example of the change of base rule in logarithms. I never understood this rule, but yeah...


They changed the base of the logarithm from x to 5, hence the "change of base" name.
Original post by Zacken
They changed the base of the logarithm from x to 5, hence the "change of base" name.


Hmm.. Yeah. I understood it in a different way but yeh xD.


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Original post by Zacken
They changed the base of the logarithm from x to 5, hence the "change of base" name.


Original post by RDKGames
Yes.


Is there an easier way of solving this equation. The first line of the picture is the equation.
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Original post by Chittesh14
Is there an easier way of solving this equation. The first line of the picture is the equation.


Yeah.

log⁡4x=log⁡2xlog⁡24=log⁡2x2\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}.

So your equation is just 32log⁡2x=2  ⟺  log⁡2x=43  ⟺  x=24/3\frac{3}{2} \log_2 x = 2 \iff \log_2 x = \frac{4}{3} \iff x = 2^{4/3}
Original post by Zacken
Yeah.

log⁡4x=log⁡2xlog⁡24=log⁡2x2\log_4 x = \frac{\log_2 x}{\log_2 4} = \frac{\log_2 x}{2}.

So your equation is just 32log⁡2x=2  ⟺  log⁡2x=43  ⟺  x=24/3\frac{3}{2} \log_2 x = 2 \iff \log_2 x = \frac{4}{3} \iff x = 2^{4/3}


Wow. Thanks lol


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Not sure why you even bothered with base 10, seems long winded, just use the example previously. You can use Zacken's method but I'm unsure if that rule is taught at C2, and if it is, you should try to use a base that will make the equation flow better, ie base 2 because it flows nicely with the 2 on the other side. Too tired to explain it mathematically :P

ImageUploadedByStudent Room1467502982.371851.jpg


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(edited 7 years ago)
Original post by RDKGames
ie base 2 because it flows nicely with the 2 on the other side. Too tired to explain it mathematically :P



No, that's not why you'd use base 2. You'd still convert all logarithms to base 2 if you had log⁡4x+log⁡2x=a\log_4 x + \log_2 x = a for any a∈Ra\in \mathbb{R}.
Original post by Zacken
No, that's not why you'd use base 2. You'd still convert all logarithms to base 2 if you had log⁡4x+log⁡2x=a\log_4 x + \log_2 x = a for any a∈Ra\in \mathbb{R}.


I didn't look into this much last night, but you can use any arbitrary base. Sure, the answer may not look as nice without base 2 but you would still get a correct answer. So surely base 2 is just to make things simpler?

Just re-read my previous comment, yeah it isn't the exact reason as it's completely independent of any logs.
(edited 7 years ago)
Original post by RDKGames
I didn't look into this much last night, but you can use any arbitrary base. Sure, the answer may not look as nice without base 2 but you would still get a correct answer. So surely base 2 is just to make things simpler?

Just re-read my previous comment, yeah it isn't the exact reason as it's completely independent of any logs.


Yes, it's to make things simpler. But it's not because "of the 2 on the other side of the equals sign" as you claim. It would be just as simple if there was aa on the other side of the equals sign.
Lol why do you lot argue over Maths :s-smilie:. I get confused as I don't know what's going on.


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Original post by Chittesh14
Lol why do you lot argue over Maths :s-smilie:. I get confused as I don't know what's going on.


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We're not arguing. I'm just getting something cleared up. If I wanted to argue I'd go to the Brexit threads. :smile:
Lol Ik you lot are not arguing. Maybe, it's just the way you lot speak lol - blunt.


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Original post by Chittesh14
Lol Ik you lot are not arguing. Maybe, it's just the way you lot speak lol - blunt.


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You are up far too early :s-smilie:
Original post by Chittesh14
Lol Ik you lot are not arguing. Maybe, it's just the way you lot speak lol - blunt.


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Good, clears things up faster. Nothing wrong with being blunt when talking about maths; you're either right or wrong. :tongue:
Original post by SeanFM
You are up far too early :s-smilie:


Work experience bro. Lol.. I wake up at 2 pm usually


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Original post by RDKGames
Good, clears things up faster. Nothing wrong with being blunt when talking about maths; you're either right or wrong. :tongue:


Fair enough.


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Original post by RDKGames
x


Original post by Zacken
x


Original post by SeanFM
x


Is my answer right in terms of explanation or have I blundered somewhere.
The answer is 3 I got that :redface:.

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