[Math] How does the base of a logarithmic function affect the graph

calculuslogarithms

$$\log_{10}(x)=y$$

$$\log_{20}(x)=y$$

I noticed that after cutting through $(1,0)$ the function $y=\log_{10}(x)$ is steeper .

$x=10^y$ , $x=20^y$

From this , the $x$ values must be the same, therefore the $y$ value on $20^y$ will be smaller .

However, now , let's compare

$$\log_{0.5}(x)=y$$

$$\log_{0.9}(y) = y$$

From the first example , we learnt that the bigger the base , the less steep the graph will be after $(1,0)$.
Why is this not the case for these $2$ functions?

$0.9$ is a bigger number than $0.5$, but why does the '$0.9$' graph continue to be steeper than the '$0.5$' graph after $(1,0)$?

Best Answer

Note that the function $0.5^x$ is "steeper" than $0.9^x$, because $0.5^x$ decreases more rapidly. Their inverse functions are negative for $x>1$, so therefore the function that the lesser function will be more steep. Also, consider the derivatives of $\log_{0.9}x$ and $\log_{0.5}x$: $(\log_{0.9}x)' = \frac{1}{x\ln{0.9}}$, and $(\log_{0.5}x)' = \frac{1}{x\ln{0.5}}$. As $\frac{1}{\ln{0.9}} \approx -9.491$, and $\frac{1}{\ln{0.5}} \approx -1.443$, we see from here that the function $\log_{0.9}x$ will decrease more intensely than $\log_{0.5}x$.

To see this more clearly, just remember that things "reverse" when the base of the logarithm becomes lesser than $1$, compared to when it was greater than $1$.

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