Algebra solutions

  1. Evaluate the following. Can you do this without calculating each factor separately?

    1. \((\log_3 9)(\log_9 81)\)

    2. \((\log_3 81)(\log_{81} 9)\)

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

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

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

  1. Write the following as a single logarithm. What conditions must \(x\) and \(y\) satisfy?

    1. \((\log_{10} 5)(\log_{5} x)\)

    2. \((\log_{10} x)(\log_{x} y)(\log_{y} 3)\)

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. Simplify the following. What conditions must \(x\) and \(y\) satisfy?

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

    2. \((\log_{x} y)(\log_{8} x)(\log_{y} 8)\)

  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 \(1\).

  1. The value of \(\log_2 3\) is approximately \(1.6\). Can you use this to approximate \(\log_4 27\)? What about \(\log_8 27?\)

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

From the definition of logarithms, we know that \(\log_{4}3\) is smaller than \(\log_{2} 3\), so we can expect the answer to be less than \(4.8\). In fact, we can say more than this, because \(\log_{4} 3< 1 < \log_{2} 3\), so we should expect the answer to be less than \(3\).

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 \(\log_{8} 27 = 3(\log_{8}2)(\log_{2}3)= \log_{2}3 \approx 1.6\).

Note that from the second part we have \(\log_{2^{3}}3^{3}= \log_{2} 3\). In general, \(\log_{a^{k}}b^{k}= \log_{a} b\) for any positive \(a\), \(b\) with \(a\neq 1\) and any power \(k\). How could you prove this?

  1. Evaluate

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

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

  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. Given that \(\log_x y=4\) and \(\log_y 2 = 9\), evaluate

    1. \((\log_x 2)(\log_y x)\)

    2. \(\dfrac{\log_x 2}{\log_y x}\)

  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 \(\dfrac{\log_{x}x}{\log_x y}\).

Therefore \[\begin{align*} \frac{\log_x 2}{\log_y x} &{}= \frac{(\log_x y)(\log_{y}2)}{\dfrac{\log_{x}x}{\log_x y}}\\ &{}= (\log_{x}y)^{2}\log_{y}2\\ &{}= 16 \times 9\\ &{}= 144 \end{align*}\]
  1. What is \((\log_a b^3)(\log_b a^3)\)?

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

  1. What is the relationship between \(\log_a b\) and \(\log_b a\)?

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?

  1. Solve \(\log_{x} 7 = \log_{7} x\).

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.