## Sunday, August 30, 2015

### Analysis Quiz 14: Challenging Mathematics Drill Quiz

Question 1: What can you say to the sum of the following two logarithm terms?

$\log_2 (3)+\log_6 (8)$

A. Their sum is less than zero.

B. Their sum is greater than zero.

In order to answer the question correctly, we must be super familiar with how the graph of $y=\log_{10} x$ behaves.

Note that for every negative $x$, $y=\log_{10} x$ is undefined.

Whereas for every $x\lt 1$, $y=\log_{10} x\lt 0$.

At $x=1$, $y=\log_{10} 1=0$.

At $x\gt 1$, $y=\log_{10} x\gt 0$.

Since the arguments for both logarithmic terms above are $3$ and $8$, which are above $1$, therefore each of them is greater than $0$ and so is their sum.

Therefore the answer for question $1$ is B.

Question 2: What can you say to the sum of the following two logarithm terms?

$\log_{0.1} (0.2)+\log_{0.2} (0.3)$

A. Their sum is less than zero.

B. Their sum is greater than zero.

As you can tell, the bases are no longer the common logarithm that has the base of $10$, we have the logarithms of the bases of $0.2$ and $0.3$ in this question.

As long as you know how to use the change-of-base formula for logarithm, then you will be able to answer this question correctly.

\begin{align*}\log_{0.1} (0.2)+\log_{0.2} (0.3)&=\dfrac{\log_{10} (0.2)}{\log_{10} (0.1)}+\dfrac{\log_{10} (0.3)}{\log_{10} (0.2)}\\&=\dfrac{\text{negative}}{\text{negative}}+\dfrac{\text{negative}}{\text{negative}}\\&\gt 0\end{align*}

Therefore, the answer for this question is B.

Question 3: What can you say to the sum of the following two logarithm terms?

$\log_{2} (0.2)+\log_{2} (0.3)$

A. Their sum is less than zero.

B. Their sum is greater than zero.

When we have already the strong foundation of the behavior of the graph of $y=\log_{10} x$, then this question is really an easy peasy one, we can say out so loud that $\log_{2} (0.2)$ and $\log_{2} (0.3)$ are both negative, thus, their sum is a negative as well.

A is hence the answer for this problem.

Question 4: What can you say to the difference of the following two logarithm terms?

$\log_{3} (0.4)-\log_{2} (0.4)$

A. The difference is less than zero.

B. The difference is greater than zero.

Note that the argument $0.4$ of these both logarithm terms is the same, it's just that they have the different bases.

Note that for $x\lt 1$, the graph of $y=\log_{3} (x)$ lies above $y=\log_{2} (x)$. That means $\log_{2} (0.4)$ is more negative than $\log_{3} (0.4)$, therefore,

\begin{align*}\log_{3} (0.4)-\log_{2} (0.4)&=\text{a negative}-\text{a more negative}\gt 0\end{align*}

Question 5: Given "$k$" is the set of positive integer. What can you say to the sum of the following two logarithm terms?

$\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))$

A. Their sum is less than zero.

B. Their sum is greater than zero.

It's more easy to look at the each term and analysis them separately:

For [MATH]\color{yellow}\bbox[5px,purple]{\log_k (\log_k (k+2))}[/MATH]:

Note that for $k$ be the positive integer, $k+2\gt 1$ and $k+2\gt k$, all those suggest $\log_k (k+2)\gt 1$.

Similarly, for $\log_k (k+2)\gt 1$, we can say [MATH]\color{yellow}\bbox[5px,purple]{\log_k (\log_k (k+2))\gt 0}[/MATH].

For [MATH]\color{yellow}\bbox[5px,green]{\log_{k+2} (\log_{k+2} (k+4))}[/MATH]:

Note that for $k$ be the positive integer, $k+2\gt 1$, $k+4\gt 1$ and $k+4\gt k+2$, all those suggest $\log_{k+2} (k+4)\gt 1$.

Similarly, for $\log_{k+2} (k+4)\gt 1$, we can say [MATH]\color{yellow}\bbox[5px,green]{\log_{k+2} (\log_{k+2} (k+4))\gt 0}[/MATH].

Thus their sum is of course greater than zero. B is hence the answer.

Question 6: Let's say the sum of the two logarithm terms in question 5 is greater than zero. What should be the greatest point of concern that might change that inequality if we are to add another logaritm term into the LHS of the left hand side of the inequality?

$\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))\gt 0$

$\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))+\log_{k+4}(\log_{k+4} (k))$

A. The new sum of three logarithm terms might be a negative.

B. The new addend is a negative term.

C. None of the above.

We can see that $k\lt k+4$, this gives $\log_{k+4} (k))\gt 1$.

This consequently makes $\log_{k+4}(\log_{k+4} (k))$ a negative.

As long as we haven't figured out if the sum of $\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))$ is greater or less than $\log_{k+4}(\log_{k+4} (k))$, we can't say for sure if $\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))+\log_{k+4}(\log_{k+4} (k))$ is greater or less than zero.

Therefore, B is the answer for this question 6.

Question 7: Question 7: What is your conclusion for question 6?

$\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))+\log_{k+4}(\log_{k+4} (k))$

A. The sum of the given three terms is less than zero.

B. The sum of the given three terms is greater than zero.

We need to ascertain if the sum of the two positive terms $\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))$ is less than or greater than $\log_{k+4}(\log_{k+4} (k))$ before drawing for any kind of conclusion.

There is a workaround for this problem, if we can rewrite the sum of the two positive term and one negative term so it takes another form to become, says, the sum of two positive terms, then their sum is surely greater than zero.

That is a great idea! We could use the change-of-base formula to turn $\log_{k+4}(\log_{k+4} (k))$ into something else that might be helpful:

\begin{align*}\log_{k+4}(\log_{k+4} (k))&=\log_{k+4}\left(\dfrac{\log_{k+2} (k)}{\log_{k+2}(k+4)}\right)\\&=\log_{k+4}(\log_{k+2} (k))-\log_{k+4}(\log_{k+2} (k+4))\\&=\log_{k+4}\left(\dfrac{\log_{k} (k)}{\log_{k}(k+2)}\right)-\left(\dfrac{\log_{k+2}(\log_{k+2} (k+4))}{\log_{k+2}(k+4)}\right)\\&=\log_{k+4}(\log_{k}(k))-\log_{k+4}(\log_{k}(k+2))-\left(\dfrac{\log_{k+2}(\log_{k+2} (k+4))}{\log_{k+2}(k+4)}\right)\\&=\log_{k+4}(1)-\log_{k+4}(\log_{k}(k+2))-\left(\dfrac{\log_{k+2}(\log_{k+2} (k+4))}{\log_{k+2}(k+4)}\right)\\&=-\log_{k+4}(\log_{k}(k+2))-\left(\dfrac{\log_{k+2}(\log_{k+2} (k+4))}{\log_{k+2}(k+4)}\right)\\&=-\dfrac{\log_{k}(\log_{k}(k+2))}{\log_{k} (k+4)}-\left(\dfrac{\log_{k+2}(\log_{k+2} (k+4))}{\log_{k+2}(k+4)}\right)\end{align*}

Now, we can replace this form of $\log_{k+4}(\log_{k+4} (k))$ into $\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))+\log_{k+4}(\log_{k+4} (k))$:

$\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))+\log_{k+4}(\log_{k+4} (k))$

$=\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))-\dfrac{\log_{k}(\log_{k}(k+2))}{\log_{k} (k+4)}-\left(\dfrac{\log_{k+2}(\log_{k+2} (k+4))}{\log_{k+2}(k+4)}\right)$

$=\log_k (\log_k (k+2))\left(1-\dfrac{1}{\log_{k} (k+4)}\right)+\log_{k+2} (\log_{k+2} (k+4))\left(1-\dfrac{1}{\log_{k+2} (k+4)}\right)$

$=\log_k (\log_k (k+2))\left(\dfrac{\log_{k} (k+4)-1}{\log_{k} (k+4)}\right)+\log_{k+2} (\log_{k+2} (k+4))\left(\dfrac{\log_{k+2} (k+4)-1}{\log_{k+2} (k+4)}\right)$

Since $\log_{k} (k+4)$ and $\log_{k+2} (k+4)$ are both greater than $1$, we can safely say it out loud that $\log_k (\log_k (k+2))+\log_{k+2} (\log_{k+2} (k+4))+\log_{k+4}(\log_{k+4} (k))\gt 0$

Question 8: Can "k" be a set of positive real number (but not strictly to integer) and the inequality that you have selected from question 7 still holds?

A. Yes

B. No.

Yes, we could do that because changing the value of $k$ but to other positive real number only changes the magnitude of the value of the inequality, without changing its sign.