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September 21, 2019, 06:24:59 am

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#### DrewN20

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« Reply #4320 on: August 13, 2019, 11:44:50 am »
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Hi Guys,
Bit confused over logs atm (in year 11)
the question says:
"Find which two integers each expression lies between"
Log2(50) and the answers say 5 & 6 can someone explain this please!

#### DrewN20

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« Reply #4321 on: August 13, 2019, 11:49:54 am »
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Pls dont worry... have since got it! : |

#### Coolmate

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« Reply #4322 on: August 23, 2019, 10:16:22 pm »
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Hello Everyone!

Would someone please be able to help me with these questions based on Logarithms, b,e, and h?(Attached) I have no idea about how to go about answering them --> and also explain a bit about natural logs compared against normal logs?

Cheers,

Coolmate
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#### RuiAce

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« Reply #4323 on: August 23, 2019, 10:31:40 pm »
+1
Hello Everyone!

Would someone please be able to help me with these questions based on Logarithms, b,e, and h?(Attached) I have no idea about how to go about answering them --> and also explain a bit about natural logs compared against normal logs?

Cheers,

Coolmate
Your calculator should have a button that can compute all of those for you. And then you just round it.

The "natural" logarithm is just a special name we give to the base $e$ logarithm, where $e$ is this fancy number (Euler's number) that behaves like $\pi$ with its weird decimals and $e\approx 2.718281828459045$.

Saying $y = \ln x$ is the exact same thing as saying $y = \log_e x$.

#### Coolmate

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« Reply #4324 on: August 23, 2019, 11:00:02 pm »
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Your calculator should have a button that can compute all of those for you. And then you just round it.

The "natural" logarithm is just a special name we give to the base $e$ logarithm, where $e$ is this fancy number (Euler's number) that behaves like $\pi$ with its weird decimals and $e\approx 2.718281828459045$.

Saying $y = \ln x$ is the exact same thing as saying $y = \log_e x$.

Hey Rui!

Thankyou! I just checked with the calculator and it does have the button!

Also, just clarifying; so ln is essentially just exactly written as loge?

Cheers,

Coolmate
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#### RuiAce

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« Reply #4325 on: August 23, 2019, 11:01:08 pm »
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Hey Rui!

Thankyou! I just checked with the calculator and it does have the button!

Also, just clarifying; so ln is essentially just exactly written as loge?

Cheers,

Coolmate
Awesome

And yep

#### Coolmate

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« Reply #4326 on: August 23, 2019, 11:04:35 pm »
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Thanks Rui for helping me! That's awesome

Cheers,

Coolmate
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#### mirakhiralla

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« Reply #4327 on: August 29, 2019, 10:47:30 pm »
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Hey sorry I need help in ii
In the Jackpot Lottery, the probability of the Jackpot prize being won in any draw is approximately 1 in 50.

i) What is the probability that the jackpot prize will be won in each of the three consecutive draws?

ii) How many consecutive draws must be made for it to be 99% certain that a Jackpot prize will have been won?

#### DrDusk

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« Reply #4328 on: August 29, 2019, 11:36:48 pm »
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Hey sorry I need help in ii
In the Jackpot Lottery, the probability of the Jackpot prize being won in any draw is approximately 1 in 50.

i) What is the probability that the jackpot prize will be won in each of the three consecutive draws?

ii) How many consecutive draws must be made for it to be 99% certain that a Jackpot prize will have been won?

$\text{Since there is a 1/50 change of winning, we can say there is a 49/50 chance of losing}$
$\text{So what we want to do is do 49/50 * 49/50 'n' number of times/draws such that the probability of losing is 0.01}$
$\therefore \bigg(\frac{49}{50}\bigg)^n = 0.01$
$\therefore nln\bigg(\frac{49}{50}\bigg) = ln(0.01)$
$\therefore n = 228\hspace{2mm}\text{draws}$
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#### RuiAce

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« Reply #4329 on: August 30, 2019, 10:14:21 am »
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Hey sorry I need help in ii
In the Jackpot Lottery, the probability of the Jackpot prize being won in any draw is approximately 1 in 50.

i) What is the probability that the jackpot prize will be won in each of the three consecutive draws?

ii) How many consecutive draws must be made for it to be 99% certain that a Jackpot prize will have been won?
Basically see above for the answer. Just want to provide a small remark for extra intuition.

The question is probably doable directly, but it would probably be messy. Because you only require the probability that it is won at least one out of the first $n$ draws, you get different results depending on if it's won exactly once, twice, three times, all the way up to $n$ times.

And in 2U maths, of course we learn that the complement is a natural way to navigate around the "at least" issue wherever possible. The complement is when you don't win it at all, which you know can only happen one possible way. (Namely, it is never won.)

So this probability will be $\left( \frac{49}{50} \right)^n$, and hence what we require is $1 - \left( \frac{49}{50} \right)^n = 0.99$. Which of course, becomes what was computed above.
-snip-
You can use \times for $\times$ and you should use \ln for $\ln$ for better LaTeX in the future

#### Kombmail

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« Reply #4330 on: September 09, 2019, 06:17:46 pm »
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does anyone know how Ln(1)= Ln(d)
becomes d=e?
-KgkG-

#### RuiAce

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« Reply #4331 on: September 09, 2019, 06:19:45 pm »
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does anyone know how Ln(1)= Ln(d)
becomes d=e?
Except it doesn’t, so somewhere in there is a mistake.

$\ln d=1$ becomes $d=e$.
« Last Edit: September 09, 2019, 06:24:23 pm by RuiAce »

#### Coolmate

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« Reply #4332 on: September 11, 2019, 09:58:09 pm »
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Hi Everyone!

I am in Year 11 revising Calculus for my prelims and was wondering if someone would be able to step through how to do the Product Rule with these questions below (attached). I am confused with 'a', 'h', and 'i'.

Coolmate
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#### RuiAce

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« Reply #4333 on: September 11, 2019, 10:04:28 pm »
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Hi Everyone!

I am in Year 11 revising Calculus for my prelims and was wondering if someone would be able to step through how to do the Product Rule with these questions below (attached). I am confused with 'a', 'h', and 'i'.

Coolmate
I'll only do i) for now. For the rest, if you have further trouble please post up any progress.
\begin{align*}
u &= x+1 &&& v &= (2x+5)^4\\
u^\prime &= 1 &&& v^\prime &= 8(2x+5)^3
\end{align*}
Note that the chain rule was required to obtain $v^\prime$. The "inner" function was $2x+5$ and the outer function was $(...)^4$.
\begin{align*}
\frac{dy}{dx} &= vu^\prime + u v^\prime\\
&= (2x+5)^4 + 8(2x+5)^3(x+1)\\
&= (2x+5)^3\left[ (2x+5) + 8(x+1) \right] \\
&= (2x+5)^3 (10x+13)
\end{align*}
« Last Edit: September 11, 2019, 10:06:06 pm by RuiAce »

#### fun_jirachi

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