Which of the two logarithms is larger: log_7 (10) or log_11 (13)

[anemone]

Without using a calculator, prove which of the following is larger:

\(\displaystyle \log_7 10\) or \(\displaystyle \log_{11} 13\)

 

[Opalg]

Without using a calculator, prove which of the following is larger:

\(\displaystyle \log_7 10\) or \(\displaystyle \log_{11} 13\)

[sp]Using logs to base 10, $\log_7 10 = \dfrac{\log10}{\log7}$ and $\log_{11}13 = \dfrac{\log13}{\log11}.$ If $$\log10\cdot \log11 > \log7\cdot \log13\qquad(*)$$ then it will follow that $\log_7 10 > \log_{11} 13$.

Since $\log10 = 1$ and $\log11>1$, the left side of (*) is greater than $1$. To see that the right side of (*) is less than $1$, notice that $7^7 = 823543 < 10^6$ and $13^7 = 62748517 < 10^8$. (In principle, you could get those results by repeated multiplication without using a calculator, though I admit that I cheated there.) Therefore $7\log7<6$ and $7\log13<8$, and so $$\log7\cdot \log13 <\frac67\cdot\frac87 = \frac{48}{49} < 1,$$ which shows that the inequality (*) holds.[/sp]

 

[tkhunny]

Without using a calculator, prove which of the following is larger:

\(\displaystyle \log_7 10\) or \(\displaystyle \log_{11} 13\)

Third Grader, non-general Solution. :-)
[sp]I want this to be WAY easier.

10/7 = 1 + ~43% (no calculator required)

13/11 = 1 + ~18% (no calculator required)

It's going to take A LOT more work to get 43% out of 7 than is required to get 18% out of 11.

No doubt, other examples would be less clear to my third grade friend.[/sp]

 

[MarkFL]

My solution:

Spoiler

In much the same way as Chris, I got to the point where I needed to show:

\(\displaystyle 1>\log(7)\log(13)=\log(10-3)\log(10+3)\)

Note: I am using the notation \(\displaystyle \log(x)=\log_{10}(x)\).

Now, obviously:

\(\displaystyle \log(10-3)<1\) and \(\displaystyle \log(10+3)>1\)

And given the fact that the rate of change of log functions varies inversely as their argument, we could choose to write:

\(\displaystyle \log(13)=1+x\)

\(\displaystyle \log(7)=1-(x+y)\)

where \(x,y>0\)

And so our inequality becomes:

\(\displaystyle 1>(1-(x+y))(1+x)=1-(y+x^2+xy)\)

or:

\(\displaystyle 0<y+x^2+xy\)

As \(x\) and \(y\) are both defined to be positive, the result follows.

 

[anemone]

Thanks to Opalg and MarkFL for participating and for the two insightful solutions!

Now, @tkhunny, I am afraid you have to walk me through your solution...

My solution:

Spoiler

First, we assume that \(\displaystyle \log_7 10>\log_{11} 13\) is true. By changing the base-7 and base-11 logarithms to base-10 logarithms, we have

\(\displaystyle \begin{align*}\frac{\log_{10} 10}{\log_{10} 7}&>\frac{\log_{10} 13}{\log_{10} 11}\\ \frac{1}{\log_{10} 7}&>\frac{\log_{10} 13}{\log_{10} 11}\\ \log_{10} 11&>\log_{10} 7\log_{10} 13\,\,\,\,\,\,\text{(since }\log_{10} x>0\,\,\, \text{for all}\,\,x>1)\end{align*}\)

We can see that \(\displaystyle \log_{10} 11>1\) as \(\displaystyle 11>10\). So, if we can prove \(\displaystyle 1>\log_{10} 7\log_{10} 13\), we are done.

Observe that by the AM-GM inequality, we have

\(\displaystyle \sqrt{\log_{10} 7\log_{10} 13}<\frac{\log_{10} 7+\log_{10} 13}{2}=\frac{\log_{10} (7(13))}{2}=\frac{\log_{10} 91}{2}<\frac{\log_{10} 100}{2}=1\\\implies 1>\log_{10} 7\log_{10} 13\)

The proof is then followed.

 

[tkhunny]

Now, @tkhunny, I am afraid you have to walk me through your solution...

I just wondered if there were a range over which we could simply ignore the logarithms.

There appears to be such a range, but the non-calculator premise may be violated when making the less vague impression.

 

[kaliprasad]

Without using a calculator, prove which of the following is larger:

\(\displaystyle \log_7 10\) or \(\displaystyle \log_{11} 13\)

Spoiler

we have $10 > 7 * 1.4$ and $13 < 11 * 1.4$
hence $log_7 \,10 > 1 + log_7 \,1.4$ and $log_{11} \,13 = 1 + log_{11} \,1.4$
so $log_7\,10 > 1 + log_7 \,1.4 > 1 + log_{11} \,1.4 > log_{11} \,13$ as $log_7 \,1.4 > log_{11} \,1.4$
hence $log_7\,10 > log_{11} \,13$