Which of these logarithms has the biggest value?

[songoku]

Homework Statement

Which one is the biggest?
a. $\log_{2015}2016$
b. $\log_{2016}2017$
c. $\log_{2017}2018$
d. $\log_{2018}2019$
e. $\log_{2019}2020$

Relevant Equations

Logarithm properties

Is there any way to answer the question without just calculating it using calculator, maybe manipulating the number using logarithm properties?

Thanks

 

[PeroK]

Can you see which one must be largest? You could prove it use log identities.

 

[songoku]

Can you see which one must be largest? You could prove it use log identities.

I can't, without using calculator

$\log_{2015}2016=\frac{\log 2016}{\log 2015}$

$\log_{2016}2017=\frac{\log 2017}{\log 2016}$

$\log_{2017}2018=\frac{\log 2018}{\log 2017}$

$\log_{2018}2019=\frac{\log 2019}{\log 2018}$

$\log_{2019}2020=\frac{\log 2020}{\log 2019}$

From option (a) to (e), both numerator and numerators become larger so I don't know about their ratio.Trying to change it into index form:
$\log_{2015}2016=a \rightarrow 2016 = 2015^{a}$

$\log_{2016}2017=b \rightarrow 2017 = 2016^{b}$

$\log_{2017}2018=c \rightarrow 2018 = 2017^{c}$

$\log_{2018}2019=d \rightarrow 2019 = 2018^{d}$

$\log_{2019}2020=e \rightarrow 2020 = 2019^{e}$

What logarithm properties do I need to use to find the order of the number?

Thanks

 

[PeroK]

I can't, without using calculator

$\log_{2015}2016=\frac{\log 2016}{\log 2015}$

$\log_{2016}2017=\frac{\log 2017}{\log 2016}$

$\log_{2017}2018=\frac{\log 2018}{\log 2017}$

$\log_{2018}2019=\frac{\log 2019}{\log 2018}$

$\log_{2019}2020=\frac{\log 2020}{\log 2019}$

From option (a) to (e), both numerator and numerators become larger so I don't know about their ratio.

What about letting $f(x) = \log_{x}(x+1) = \frac{\log (x+1)}{\log x}$ and showing that $f(x)$ is a decreasing function?

 

[songoku]

What about letting $f(x) = \log_{x}(x+1) = \frac{\log (x+1)}{\log x}$ and showing that ##f(x) is a decreasing function?

This is brilliant (I will never be able to think towards this direction).

Thank you very much PeroK