Changing bases

Using the generalised version of the result in the warm-up,

1. $(\log_3 9)(\log_9 81)=\log_{3}81=4$

2. $(\log_3 81)(\log_{81} 9)= \log_{3}9=2$.

Using the same idea as in question 1, (a) $(\log_{10} 5)(\log_{5} x)=\log_{10} x$.

For part (b), one approach is

$$(\log_{10} x)(\log_{x} y)(\log_{y} 3) = (\log_{10} y)(\log_{y} 3) = \log_{10}3$$

Since $x$ and $y$ both occur as bases of logarithms in these expressions, $x$ and $y$ must both be positive, but can’t be $1$.

1. $(\log_{x} y)(\log_{y} 10)(\log_{8} x)=(\log_{x} 10)(\log_{8} x)= (\log_{8} x)(\log_{x} 10)= \log_{8}10$.

2. $(\log_{x} y)(\log_{8} x)(\log_{y} 8)= (\log_{8} x)(\log_{x} y)(\log_{y} 8)= (\log_{8} y)(\log_{y} 8) = \log_{8}8=1$.

Again, since $x$ and $y$ both occur as bases of logarithms in these expressions, $x$ and $y$ must both be positive, but can’t be $1$.

If we start by using the law for manipulating logs of powers, we get $\log_{4} 27 = 3 \log_{4} 3$.

From the warm-up we know that $\log_{4} 3 = \frac{1}{2}\log_{2} 3$. Therefore $\log_{4} 27 = 3 (\log_{4} 2)(\log_{2}3) \approx 3 \times \frac{1}{2}\times 1.6 = 2.4$.

We know that $\log_{8} 27$ will be smaller than $\log_4 27$. By manipulating powers and changing the base, we get $\log_{8} 27 = 3(\log_{8}2)(\log_{2}3)= \log_{2}3 \approx 1.6$.

1. $\dfrac{\log_{2} 5}{\log_{2} 25}= \dfrac{\log_{2} 5}{(\log_{2}5)(\log_{5}25)}= \dfrac{1}{2}$

2. $\dfrac{\log_{10} 5}{\log_{10} 3}=\dfrac{(\log_{10} 3)(\log_{3}5)}{\log_{10} 3}=\log_{3}5$

Alternatively, we could use the fraction form of the general result in the warm-up.

1. $(\log_x 2)(\log_y x)= (\log_y x)(\log_x 2)= \log_{y} 2 = 9$.

2. Using ideas from question 5, or the two forms for changing base given in the warm-up, the numerator can be rewritten as $(\log_x y)(\log_{y}2)$ and the denominator as $\dfrac{\log_{x}x}{\log_x y}$.

Using laws for manipulating logs of powers, $$(\log_a b^3)(\log_b a^3)=9(\log_a b)(\log_b a) = 9 \log_{a} a = 9.$$

In questions 3 and 7 we had something of the form $(\log_{a} b)(\log_{b} a) = \log_{a} a$, so $(\log_{a} b)(\log_{b} a)=1$, or equivalently $\log_{a} b = \dfrac{1}{\log_{b}a}$. Is there any danger of dividing by $0$ here?

Using the relationship in question 8, solving $\log_{x} 7 = \log_{7} x$ is equivalent to solving $(\log_{7}x)^{2} = 1$. Therefore we need to solve $\log_{7} x = 1$ and $\log_{7} x = -1$, which give us $x=7$ and $x=\dfrac{1}{7}$ as solutions.