Estimating Logarithms

Algebra Level 5

Let $x$ be a real number such that $0.170 < \log_x 2 < 0.171$ and $0.270 < \log_x 3 < 0.271$ . Determine the integer that is closest in value to $\log_x 7^{100}$ .

The answer is 48.

4 solutions

Ayush Verma
Apr 7, 2016

$\log _{ x }{ 2=\cfrac { \ln { 2 } }{ \ln { x } } } \& \quad \log _{ x }{ 3 } =\cfrac { \ln { 3 } }{ \ln { x } } \\ \\ \therefore \quad max\left\{ \cfrac { 0.170 }{ \ln { 2 } } ,\cfrac { 0.270 }{ \ln { 3 } } \right\} \quad <\cfrac { 1 }{ \ln { x } } <\quad min\left\{ \cfrac { 0.171 }{ \ln { 2 } } ,\cfrac { 0.271 }{ \ln { 3 } } \right\} \\ \\ \Rightarrow \cfrac { 0.270 }{ \ln { 3 } } <\cfrac { 1 }{ \ln { x } } <\cfrac { 0.271 }{ \ln { 3 } } \\ \\ \Rightarrow 0.270\log _{ 3 }{ { 7 }^{ 100 } } \quad <\log _{ x }{ { 7 }^{ 100 } } <\quad 0.271\log _{ 3 }{ { 7 }^{ 100 } } \\ \\ \Rightarrow 27\log _{ 3 }{ 7 } \quad <\log _{ x }{ { 7 }^{ 100 } } <\quad 27.1\log _{ 3 }{ 7 } \\ \\ \Rightarrow 47.82<\log _{ x }{ { 7 }^{ 100 } } <48\\ \\ Ans=48$

i like your solution because it's neat! how did you get from step 3 to step 4? thanks!

Terrell Bombb - 4 years, 6 months ago

multiplied all by ln7^100

Ayush Verma - 4 years, 4 months ago

Clarence Chew
May 20, 2014

0.170<logx(2)<0.171 0.270<logx(3)<0.271 Adding, we get: 0.440<logx(6)<0.442 0.510<logx(8)<0.513 Thus 0.440<logx(7) 0.440 100<logx(7^100)<0.513 100 44.0<logx(7^100)<51.3. More work needs to be done. We thus look at 48<7^2. 0.950<logx(48)<logx(49) 47.5<logx(7^100) (multiplying by 50). More work needs to be done. Note 7^5<2^3+3^7 (computable), resulting in logx(7^100)<48.2. We are done. So we are sure that the answer is 48.

Comments and replies:

Calvin:

I agree with you up till $0.950 = 4 \times \log 2 + \log 3 = \log 48 < \log 49$ . I believe your next line should say $7^5 < 2^3 \times 3^7$ , instead of a +. Can you explain how you chose these numbers? Can you give an algorithm where it's not randomly trying our luck?

Calvin Lin Staff - 7 years ago

Sreejato Bhattacharya
May 20, 2014

Multiplying the given inequalities, we get as following.

$0.170\times 0.270 < \log_x 2 \times \log_x 5 < 0.171\times 0.271$ . Or, $0.0459 < \log_x 7 < 0.046341$ Or, $4.59 < \log_x 7^{100} < 4.6341$ (multiplying the sides by 100) So, $\log_x 7^{100}$ lies between 4.59 and 4.6341, which implies that the closest integer to $\log_x 7^{100}$ is 5.

Note:- $\log_x y$ here denotes log of y to base x.

[Calvin - Slight Latex edits done]

Comments and replies:

Calvin:

There are the following errors in this 1. The challenge states that $0.270 < \log_x 3 < 0.271$ , but you changed it to $\log_x 5$ .
2. $\log_x 2 \times \log_x 5 \neq \log_x 7$ . Review the rules of logarithms. It says that $\log a + \log b = \log ab$ .

Note: Since $\log_x 3 < \log_x 7$ , mulyiplying by 100 gives that $27 < \log_x 3^{100} < \log_x 7^{100}$ .

Calvin Lin Staff - 7 years ago

Hkapur 97
May 20, 2014

$0.170 < \log_x 2 \Rightarrow 0.170 < \frac{\log 2}{\log x} \Rightarrow 0.170 \log x < 0.301 \qquad 1)$

$0.270 < \log_x 3 \Rightarrow 0.270 < \frac{\log 3}{\log x} \Rightarrow 0.270 \log x < \log 3 \qquad 2) \\ 0.171 > \frac{\log 2}{\log x} \Rightarrow 0.171 > \frac {\log 2}{\log x} \Rightarrow 0.171 \log x > 0.301 \qquad 3)$

$0.271 > \log_x 3 \Rightarrow 0.271 > \frac{\log 3}{\log x} \Rightarrow 0.270 \log x > 0.477 \qquad 4)$

Subtracting $1)$ from $2)$ , we get $0.1 \log x < 0.176 \Rightarrow \log x 0.176 \Rightarrow \log x > 1.76$

So by the ultra-squeeze theorum (I just invented that), $\log x = 1.76$

$\Rightarrow \log_x 7^{100} = 100 \times \log 7 \div \log x = 100 * 0.845 \div 1.76 \approx 48$

[Several edits - Calvin]

Comments and replies:

hkapur97:

oh no! the latex misinterpreted it!

Second line -

$0.270 < \log_x 3 \Rightarrow 0.270 < \frac{\log 3}{\log x} \Rightarrow 0.270 \log x \frac{\log 2}{\log x} \Rightarrow 0.171 \log x > 0.301 \dots 3)$

Calvin:

I've made several edits to your comment in trying to understand it. Let me know if it's accurate now. Typing LaTex can be difficult if you are not able to edit it. I usually type it elsewhere to check for accuracy, before posting it. (I also have the luxury of editing the comments/posts.)

1) We do not know / are given that $\log 2 \approx 0.301$ , $\log 3 \approx 0.477$ or $\log 7 \approx 0.845$ . Remember that no calculators are allowed. If you knew this beforehand, you can directly use change on base on the inequalities.

2) You can not use $\log 2 0.301$ in 3). One of these statements is wrong.

3) You cannot subtract 1) from 2) in this way. Review your basic rules of inequalities. For example, we know that $0 < 1, -5 < 1$ , and it is not true that $0-(-5) < 1-1$ . In more generality, if $A< B , C < D \Rightarrow A-C < B-D$ , then it must also be true that $\Rightarrow C-A < D-B \Rightarrow B-D < A-C$ , which is a contradiction. This is a common mistake made when dealing with inequalities. What is the correct conclusion?

4) Assuming your calculations are correct, your 'ultra squeeze' means that $\log x$ does not exist. The conclusion of $A < B$ and $B < A$ is that you have a contradiction, and there must have been a mistake somewhere (possibly in your assumptions, or even in the question).

5) $x$ can actually take a (small) range of values, so we cannot calculate an exact value of $\log x$ .

hkapur97:

\(.170 < \log_x 2 < 0.171 \text{ and } 0.270 < \log_x 3 < 0.271\\

\Rightarrow 0.170 < \frac{\log 2}{\log x} < 0.171 \\

\Rightarrow 0.270 < \frac{\log 3}{\log x} < 0.271 \\

\Rightarrow \frac{1}{\log x} \in \bigg(\frac{0.170}{0.301} , \frac{0.171}{0.301}\bigg) \cup \bigg(\frac{0.270}{0.471}, \frac{0.271}{0.477} \bigg) \\

\Rightarrow \frac{100 \times \log 7}{\log x} \in \bigg(\frac{100 \times 0.845 \times0.170}{0.301} , \frac{100 \times 0.845 \times 0.171}{0.301}\bigg) \cup \bigg(\frac{100 \times 0.845 \times 0.270}{0.471}, \frac{100 \times 0.845 \times 0.271}{0.477} \bigg) \\

\Rightarrow \log_x7^{100} \in (47.724, 48) \cup (48, 48.439) \)

I believe this is sufficient to conclude the answer is 48. $\log 2, \log 3, \log 7$ were known beforehand. I often find it useful to have these 3 memorized.

Calvin:

Hkapur, Please read my earlier comment. I have the same objections (and then some more)

(Same as before) We are not given the values of $\log_{10} 2, \log_{10} 3$ , and hence may not use them directly. However, if you wish, you can explain how to calculate those values to that degree of accuracy (without using a calculator to find their values).
(Same as before) I can assure you that $\log_{10} 2 \neq 0.301$ , as it is not rational. You may not assume that $\log_{10} 2 > 0.301$ and $\log_{10} 2 < 0.301$ to obtain both sides of the inequality. Likewise, $\log_{10} 7 \neq 0.845$ . (Somewhat related) Is the 0.471 a typo? I'm not sure where you're getting it from. It appears throughout.
Review your mathematical logic. You should be using the intersection, as opposed to the union. While the statement is still true for the union, the stricter set would be the intersection. Assuming your calculations are correct (which I'm not checking because I already voiced concerns that they are wrong), then you have no integer value. For clarity, there is $Y$ that satisfies both $-1 < Y < 1$ and $1 < Y < 3$ . You cannot take the union and conclude that $-1 < Y < 3$ , hence there are numerous solutions.
Your approach will no longer work for numbers for which you have not memorized the values of. I can easily change the values. For example, do you know what $\log 1.9$ is?
[Side note] Go easy on the \ in the Latex. I had to clean up several of them in order for your code to display correctly. It also helps to break up the code naturally, instead of trying to type it all in one chunk. Your edit came in just as I posted this, so I deleted your repost.

hkapur97:

$\\ 0.170 < \log_x 2 < 0.171 \text{ and } 0.270 < \log_x 3 < 0.271 \\ \\ \Rightarrow 0.170 < \frac{\log 2}{\log x} < 0.171 \text{ and } 0.270 < \frac{\log 3}{\log x} < 0.271 \\ \\ \Rightarrow \frac{0.170}{\log 2} < \frac{1}{\log x} < \frac{0.171}{\log 2} \text{ and } \frac{0.270}{\log 3} < \frac{1}{\log x} < \frac{0.271}{\log 3} \\ \\ \Rightarrow \frac{100 \times \log 7 \times 0.170 }{\log 2} < \frac{100 \times \log 7 }{\log x} < \frac{100 \times \log 7 \times 0.171}{\log 2} \text{ and } \\ \\ \frac{100 \times \log 7 \times 0.270}{\log 3} < \frac{100 \times \log 7}{\log x} < \frac{100 \times \log 7 \times 0.271}{\log 3} \\ \\ \Rightarrow \frac{17 \times \log 7}{\log 2} < \log_x 7 ^{100} < \frac{17.1 \times \log 7}{\log 2} \text{ and } \frac{27 \times \log 7}{\log 3} < \log_x 7 ^{100} < \frac{27.1 \times \log 7}{\log 3} \\$

Do you agree with me till here?

Calvin:

Yes. You did none of 1, 2 3 or 4 that I raised. It will be fine as long as you do not try and substitute a fixed value for any of the logarithms.

Calvin Lin Staff - 7 years ago

Estimating Logarithms

The answer is 48.

4 solutions

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