Logarithm problem

[jonnburton]

Hello everybody,

I have been working on this logarithm problem and don't really understand how to deal with part of it. Could anyone point me in the right direction?

A curce, C, has equation \(\displaystyle y= x-2-2log_{10}x\), where x>0.

The vertical lines, x =6 and x= 7 meet the curve at points R and S, respectively.

a) show that the y coordinate of R is \(\displaystyle 4-log_{10}36\)


\(\displaystyle y =6-2-2log_{10}6 = 4-log_{10}6^2 = 4-log_{10}36 \)

Following the same procedure, the y coordinate of point S is

\(\displaystyle y =7-2-2log_{10}7 = 5-log_{10}7^2 = 5-log_{10}49 \)


It is this next bit where I am having trouble:

b. The area of the trapezium bounded by the lines RS, x = 6, x=7 and the x-axis is A square units. Show that

\(\displaystyle A = \frac{p}{2} -log_{10}q\), stating the values of the positive integers p and q.


The following is how I have approached this

Area of trapezium is \(\displaystyle \frac{1}{2}(a+b)h\)

\(\displaystyle \frac{1}{2}(4-log_{10}36 + 5-log_{10}49) *1\)


\(\displaystyle \frac{1}{2}(4log_{10}10-log_{10}36 + 5log_{10}10-log_{10}49) \)

\(\displaystyle \frac{1}{2}(log_{10}10^4-log_{10}36 + log_{10}10^5-log_{10}49) \)

\(\displaystyle \frac{1}{2}(log_{10}(\frac{10^4}{36}) + log_{10}(\frac{10^5}{39})) \)

\(\displaystyle \frac{1}{2}(log_{10}(\frac{10^9}{1404})) \)


The problem is this isn't leading closer to the equation for the area given, \(\displaystyle A = \frac{p}{2} -log_{10}q\). Furthermore, the answer says that p=9 and q=42 and I'm not even close there, either.

I would be very grateful for any help with this!

 

[HallsofIvy]

Hello everybody,

I have been working on this logarithm problem and don't really understand how to deal with part of it. Could anyone point me in the right direction?

A curce, C, has equation \(\displaystyle y= x-2-2log_{10}x\), where x>0.

The vertical lines, x =6 and x= 7 meet the curve at points R and S, respectively.

a) show that the y coordinate of R is \(\displaystyle 4-log_{10}36\)


\(\displaystyle y =6-2-2log_{10}6 = 4-log_{10}6^2 = 4-log_{10}36 \)

Following the same procedure, the y coordinate of point S is

\(\displaystyle y =7-2-2log_{10}7 = 5-log_{10}7^2 = 5-log_{10}49 \)


It is this next bit where I am having trouble:

b. The area of the trapezium bounded by the lines RS, x = 6, x=7 and the x-axis is A square units. Show that

\(\displaystyle A = \frac{p}{2} -log_{10}q\), stating the values of the positive integers p and q.


The following is how I have approached this

Area of trapezium is \(\displaystyle \frac{1}{2}(a+b)h\)

\(\displaystyle \frac{1}{2}(4-log_{10}36 + 5-log_{10}49) *1\)


\(\displaystyle \frac{1}{2}(4log_{10}10-log_{10}36 + 5log_{10}10-log_{10}49) \)

Your mistake is writing 4 as \(\displaystyle 4 log_{10}10\) and 5 as \(\displaystyle 5 log_{10} 10\)
just leave them as 4+ 5= 9.

\(\displaystyle \frac{1}{2}(log_{10}10^4-log_{10}36 + log_{10}10^5-log_{10}49) \)

\(\displaystyle \frac{1}{2}(9- log_{10}36- log_{10}49)= \frac{9}{2}+ \frac{log_{10}\frac{36}{49}}{2}=\frac{9}{2}+ log_{10}\left(\frac{36}{49}\right)^{1/2}= \frac{9}{2}+ log_{10}\frac{6}{7}\)
Now, what are p and q?

\(\displaystyle \frac{1}{2}(log_{10}(\frac{10^4}{36}) + log_{10}(\frac{10^5}{39})) \)

\(\displaystyle \frac{1}{2}(log_{10}(\frac{10^9}{1404})) \)


The problem is this isn't leading closer to the equation for the area given, \(\displaystyle A = \frac{p}{2} -log_{10}q\). Furthermore, the answer says that p=9 and q=42 and I'm not even close there, either.

I would be very grateful for any help with this!

 

[jonnburton]

OK, thanks for that information - I am going to have to come back to this this evening when I'll have a chance to go through it again.

 

[Deleted member 4993]

Your mistake is writing 4 as \(\displaystyle 4 log_{10}10\) and 5 as \(\displaystyle 5 log_{10} 10\)
just leave them as 4+ 5= 9.


\(\displaystyle \frac{1}{2}(9- log_{10}36- log_{10}49)= \)

\(\displaystyle \dfrac{9}{2} - \dfrac{log_{10}(36*49)}{2} = \dfrac{9}{2} - log_{10}(6*7)\)

\(\displaystyle \frac{9}{2}+ \frac{log_{10}\frac{36}{49}}{2}=\frac{9}{2}+ log_{10}\left(\frac{36}{49}\right)^{1/2}= \frac{9}{2}+ log_{10}\frac{6}{7}\)
Now, what are p and q?

.

 

[JeffM]

Hello everybody,

I have been working on this logarithm problem and don't really understand how to deal with part of it. Could anyone point me in the right direction?

A curce, C, has equation \(\displaystyle y= x-2-2log_{10}x\), where x>0.

The vertical lines, x =6 and x= 7 meet the curve at points R and S, respectively.

a) show that the y coordinate of R is \(\displaystyle 4-log_{10}36\)


\(\displaystyle y =6-2-2log_{10}6 = 4-log_{10}6^2 = 4-log_{10}36 \)

Following the same procedure, the y coordinate of point S is

\(\displaystyle y =7-2-2log_{10}7 = 5-log_{10}7^2 = 5-log_{10}49 \)


It is this next bit where I am having trouble:

b. The area of the trapezium bounded by the lines RS, x = 6, x=7 and the x-axis is A square units. Show that

\(\displaystyle A = \frac{p}{2} -log_{10}q\), stating the values of the positive integers p and q.


The following is how I have approached this

Area of trapezium is \(\displaystyle \frac{1}{2}(a+b)h\)

\(\displaystyle \frac{1}{2}(4-log_{10}36 + 5-log_{10}49) *1\)


\(\displaystyle \frac{1}{2}(4log_{10}10-log_{10}36 + 5log_{10}10-log_{10}49) \)

\(\displaystyle \frac{1}{2}(log_{10}10^4-log_{10}36 + log_{10}10^5-log_{10}49) \)

\(\displaystyle \frac{1}{2}(log_{10}(\frac{10^4}{36}) + log_{10}(\frac{10^5}{39})) \)WHOA Nellie. How did 49 become 39.

\(\displaystyle \frac{1}{2}(log_{10}(\frac{10^9}{1404})) \) 36 * 39 = 1404, but 36 * 49 = 1764


The problem is this isn't leading closer to the equation for the area given, \(\displaystyle A = \frac{p}{2} -log_{10}q\). Furthermore, the answer says that p=9 and q=42 and I'm not even close there, either.

I would be very grateful for any help with this!

First of all, HOI is quite right about your method. You are asked to derive an answer in the form of a rational number minus a logarithm. When you convert all integers to logs, you are going away from your goal rather than toward it. This is a difficult problem because you are asked to give an answer in a particular form. So it is not just a question of getting a correct answer, but a correct answer in a particular form. So, all the time you are solving, you have to remember the form in which the answer must be expressed.

Nevertheless, your method, however roundabout, will eventually get you where you need to go if you do not make an arithmetical mistake. You just gave up too soon.

\(\displaystyle \frac{1}{2}(log_{10}(\frac{10^9}{1764})) = log_{10}\left\{\left(\dfrac{10^9}{1764}\right)^{(1/2)}\right\} = log_{10}\left(\dfrac{(10^9)^{(1/2)}}{(1764)^{(1/2)}}\right) = log_{10}\dfrac{10^{(9/2)}}{\sqrt{1764}} = log_{10}\left(\dfrac{10^{(9/2)}}{42}\right).\)

If you had got here, which you might have had you not made a careless mistake, and then thought about the form required for the answer, you would have noticed that the log of the numerator of your fraction turns into a rational number.

\(\displaystyle log_{10}\left(\dfrac{10^{(9/2)}}{42}\right) = log_{10}\left(10^{(9/2)}\right) - log_{10}(42) = \dfrac{9}{2} - log_{10}(42).\)

Now this is a lot more work than is required if you remember what you are trying to achieve.

\(\displaystyle \dfrac{1}{2}(4-log_{10}36 + 5-log_{10}49) *1 = \dfrac{1}{2} * \{9 - log(36) - log(49)\} = \dfrac{9}{2} - \dfrac{1}{2}\left(log_{10}(6^2) + log_{10}(7^2)\right) = \dfrac{9}{2} - \{log_{10}(\sqrt{6^2}) + log_{10}(\sqrt{7^2})\} =\)

\(\displaystyle \dfrac{9}{2} - \{log_{10}(6) + log_{10}(7)\} = \dfrac{9}{2} - log_{10}(6 * 7) = \dfrac{9}{2} - log_{10}(42).\)

 

[jonnburton]

Many thanks indeed for all of your posts and help, I do appreciate it.

I am beginning to see the picture now. I just wondered whether someone could answer one doubt I have. I am able to follow the long-winded method shown by JeffM the most clearly. But in the easier way there is one thing I don't understand:


I don't understand why this \(\displaystyle \frac{1}{2}(9-log_{10}36-log_{10}49)\) becomes this: \(\displaystyle \frac{9}{2}-\frac{log_{10}(36*49)}{2}\) and not \(\displaystyle \frac{9}{2}-\frac{log_{10}\frac{36}{49}}{2}\)


- what I don't see is why what was \(\displaystyle log_{10}36-log{10}49\) becomes \(\displaystyle log_{10}36 + log_{10}49\)

I can see that everything has not been multiplied by -1, because the \(\displaystyle \frac{9}{2}\) remains positive.

It must be something basic that I'm overlooking but can't see at all what that is!

 

[DrPhil]

Many thanks indeed for all of your posts and help, I do appreciate it.

I am beginning to see the picture now. I just wondered whether someone could answer one doubt I have. I am able to follow the long-winded method shown by JeffM the most clearly. But in the easier way there is one thing I don't understand:


I don't understand why this \(\displaystyle \frac{1}{2}(9-log_{10}36-log_{10}49)\) becomes this: \(\displaystyle \frac{9}{2}-\frac{log_{10}(36*49)}{2}\) and not \(\displaystyle \frac{9}{2}-\frac{log_{10}\frac{36}{49}}{2}\)


- what I don't see is why what was \(\displaystyle log_{10}36-log{10}49\) becomes \(\displaystyle log_{10}36 + log_{10}49\)

I can see that everything has not been multiplied by -1, because the \(\displaystyle \frac{9}{2}\) remains positive.

It must be something basic that I'm overlooking but can't see at all what that is!

Both of the logarithms have the same sign, and they are grouped together with the minus taken outside:

\(\displaystyle \frac{1}{2}(9-log_{10}36-log_{10}49) = \frac{1}{2}(9-[log_{10}36+log_{10}49])\)

The rest of the steps follow from that, with a minus sign in front of the combined logarithm.

 

[jonnburton]

Ah, OK. Thanks for that, DrPhil!