### Exponentials & Logarithms

Package of problems

## Graph solutions

This is the graph of $y=\log_{5} x$

1. $\log_{25}x$
2. $\log_{\sqrt{5}}x$
3. $\log_{\frac{1}{5}}x$
4. $\log_{10} x$
5. $\log_{x} 5$

Try to make your graphs as accurate as you can, but use the graph of $y=\log_{5}x$ to help you, rather than a calculator.

Here are graphs 1 and 2 drawn using values from the $y=\log_{5}x$. The relationships between each graph and the graph of $y=\log_{5}x$ are described below.

1. $\log_{25} x = (\log_{25} 5)(\log_{5} x) = \frac{1}{2}\log_{5} x$, so the graph of $y = \log_{25} x$ is a stretch of the graph $y=\log_{5}x$ parallel to the $y$-axis and scale factor $\frac{1}{2}$.

2. $\log_{\sqrt{5}}x= (\log_{\sqrt{5}} 5)(\log_{5} x)=2\log_{5} x$, so the graph of $y=\log_{\sqrt{5}}x$ is a stretch of the graph $y=\log_{5}x$ parallel to the $y$-axis and scale factor $2$.

Thinking about the definition of logarithms, as we did in the warm-up, these results make sense. For example, we’d expect the power that we’d raise $5$ to in order to get a given value of $x$ to be twice as much as the power we’d need to raise $25$ to.

1. As $\dfrac{1}{5} = 5^{-1}$, the power that we’d raise $\dfrac{1}{5}$ to in order to get a given value of $x$ is $-1$ times the power we’d need to raise $5$ to. Alternatively, using the idea from the warm-up, $\log_{\frac{1}{5}}x=(\log_{\frac{1}{5}} 5)(\log_{5}x)=-\log_{5}x$ and so the graph of $y = \log_{\frac{1}{5}}x$ is a reflection in the $x$-axis of the graph of $y=\log_{5}x$.

1. More work is needed for the graph of $y=\log_{10} x$. We know that $\log_{10}x= \dfrac{\log_{5} x}{\log_{5} 10}$, so we need to divide by $\log_{5}10$ to get the graph of $y=\log_{10}x$. The given graph doesn’t go up as far as $x=10$, but $\log_{5}10 = \log_{5}5+ \log_{5}2 = 1+ \log_{5}2$. From the graph, $\log_{5} 2 \approx 0.42$. Now, $1 \div 1.42 = \frac{100}{142}=\frac{50}{71}\approx\frac{49}{70} = 0.7$ so the graph of $y=\log_{10}x$ is a stretch of $y=\log_{5}x$ parallel to the $y$-axis, scale factor $0.7$.

1. For $x>0$ with $x\neq 1$, ($\log_{5}x)(\log_{x}5)= \log_{5}5=1$, so $\log_{x}5$ is the reciprocal of $y=\log_{5}x$. We can use this to plot particular points to get an idea of the shape of the graph $y=\log_{x}5$.

Alternatively, we can use the behaviour of the function $\log_{5} x$ to sketch the behaviour. For $x>1$, as $x$ increases, $\log_{5}x$ increases and is positive, so $\log_{x}5$ decreases and is positive. For $x$ close to $1$, $\log_{5} x$ is close to $0$, and the graph of $y=\log_{x}5$ has an asymptote at $x=1$. For $0< x <1$, $\log_{5}x$ is negative. It increases to $0$ as $x$ increases to $1$, and tends to $-\infty$ as $x$ decreases to $0$. The graph of $y=\log_{x}5$ gets steeper and closer to the $y$-axis as $x$ decreases to $0$, but it doesn’t actually meet the $y$-axis because $\log_{x}5$ isn’t defined for $x=0$.