C2 graph/logs question

Hey guys,

I'm a bit stuck with part 3 of this question, rearranging the equations to get a half in the equation. I worked out the answer to the first half of using a different method but I can't quite get the method they're using to get the answer to the y coordinate of p

x = -0.32

please post your workings

...

Okay well I put the two equations equal to each other straight away and took logs of both sides:

(1/3)^x = 2(3^x)

Xlog1/3 = log 2(3^x)

xlog(1/3) = log2 + xlog3

Xlog1/3 - xlog3 = log2

X(log1/3 - log3) = log2

X= log2/log1/9

X= -0.32


Posted from TSR Mobile

If you've found $x$, then you can just plug that value right back any of your two equations or $y = 2(3^x)$ and they'll both give you the y-coordinate out.

It's kind of like how, if you have a line $y = 2x +3$ and $y = x + 4$ then they intersect at $2x + 3 = x + 4 \iff x = 1$, which is the x-coordinate of the intersection, but also the y-coordinate is given by plugging that into $y = 2x + 3 = 2(1) +3 = 5$ or $y = x + 4 = 1 +4 = 5$.

If you've found $x$, then you can just plug that value right back any of your two equations or $y = 2(3^x)$ and they'll both give you the y-coordinate out.

It's kind of like how, if you have a line $y = 2x +3$ and $y = x + 4$ then they intersect at $2x + 3 = x + 4 \iff x = 1$, which is the x-coordinate of the intersection, but also the y-coordinate is given by plugging that into $y = 2x + 3 = 2(1) +3 = 5$ or $y = x + 4 = 1 +4 = 5$.

Zacken has already started helping you
all the best

Thanks for this, I knew that but x was a very long decimal and subbing it in to get y didn't give me a nice answer of root 2, it was a few decimal places out so I didn't think I'd get the marks for doing that here? I'm not sure though


Posted from TSR Mobile

You're not meant to use the decimal answer, you should use the exact value instead:

So you know that $x = \displaystyle \frac{\log 2}{\log \frac{1}{9}}$, then:

How can you work out $3^x$ exactly? Use the fact that $3^{\log_3 m} =m$

Does that help?

Ohh yes it does! Still would never have thought to do that in exam - but what about with this method:

So if (1/3)^x = (1^x/3^x) = 2(3^x)

1^x = 2 x (3^2x)

1^x / 2 = 3^2x ?

What happens to the 1^x ?




Posted from TSR Mobile

1^x = 1 for all x.

Anyone know which paper this is from?

it's one of the OCR Solomon papers, H I think


Posted from TSR Mobile