Question

How do you prove `log_7 10 > log_11 13` ?

Answer 1

Please see below.

Explanation:

Let us convert both of them to common base `10`, then

`log_7 10=log10/log7=1/0.8451=1.1833`

and `log_11 13=log13/log11=1.1139/1.0414=1.0696`

Hence `log_7 10 > log_11 13`

Note; When we use base as `10`, we do not mention base.

Answer 2

See below.

Explanation:

Considering

`{(y_1=log_7 10),(y_2=log_11 13):} rArr {(7^(y_1)=10),(11^(y_2)=13):}`

with `y_1 > 1` and `y_2 > 1` because `7 < 10` and `11 < 13` respectively.

and

`11^(y_2)/7^(y_1) = 13/10 rArr (11/7)^(y_1) 11^(y_2-y_1) = 13/10`

or

`(((11/7)^(y_1))/(13/10))11^(y_2) = 11^(y_1)`

Now as we can easily verify

`11/7 > 13/10` and if `y_1 > 1` we have

`(((11/7)^(y_1))/(13/10)) = lambda > 1`

Now if

`lambda 11^(y_2) = 11^(y_1)` with `lambda > 1` as a consequence we have

`11^(y_2) < 11^(y_1) rArr y_1 > y_2`

or

`log_7 10 > log_11 13`

Answer 3

A solution using elementary methods...

Explanation:

This is a little lengthy, but let's use some elementary methods which do not need a calculator:

First let's find the first few powers of `13`:

`13^1 = 13`

`13^2 = 169`

`13^3 = 2197`

`13^4 = 28561`

`13^5 = 371293 < 10^6`

Taking logs of this last inequality, we find:

`5 log(13) < 6`

and hence:

`log(13) < 6/5`

We will use this later.

Note that:

`10 log(2) = log(2^10) = log(1024) > log(10^3) = 3`

Hence:

`log(2) > 0.3`

Also:

`2 log(7) + log(2) = log(2*7^2) = log(98) < log(10^2) = 2`

Hence:

`log(7) < 1/2(2-log(2)) < 1/2(2-0.3) = 0.85`

Also:

`log(13)+1 = log(13)+log(10) = log(130) > log(128) = log(2^7) = 7 log(2) > 7*0.3 = 2.1`

So:

`log(13) > 1.1`

Then:

`3 log(11) = log(11^3) = log(1331) > log(1300) = log(13)+log(10^2) = 2+log(13) > 2+1.1 = 3.1`

So:

`log(11) > 3.1/3`

So:

`log(11)/log(7) > 3.1/3*1/0.85 = 62/51 > 6/5 > log(13)`

Hence:

`log(10)/log(7) = 1/log(7) > log(13)/log(11)`

Using the change of base formula for logarithms, that is:

`log_7 10 > log_11 13`