Express $\log 162$ and $\log 1.35$ in terms of $a = \log 24$ and $b = \log 18$

Question

Here's a question in my textbook:

Express $\log 162$ and $\log 1.35$ in terms of $a$ and $b$ where $\log 24 = a$ and $\log 18 = b$.

I know in order to express $\log 162$ in the form of $xa + xb$, the equation $24^x \times 18^b = 162$ must be true. But I couldn't find such combination of $x$ and $y$.

The answer in the textbook is $\log 162 = \frac{11b - a}{5}$ and $\log 1.35 = 2b - a - 1$, but evaluating the results in backwards suggests that the process for $\log 162$ involved very large numbers.

Any help is much appreciated!

Answer 1

You know that: \begin{align*} 162 & = 2 \times 3^4 \\ 24 & = 2^3 \times 3 \\ 18 & = 2 \times 3^2 \\ \end{align*} Therefore $162 = 24^x 18^y$ is equivalent to: $$2^{3x+y}\times3^{x+2y} = 2\times 3^4$$ By uniqueness of prime decomposition, this is equivalent to: $$\left\{ \begin{array}{l} 3x+y = 1 &\\ x + 2y = 4 \end{array} \right.$$ This system has a unique solution $(x, y) = (\frac{-2}{5}, \frac{11}{5})$.

You were afraid of this process involving huge numbers. At this point if I wanted to check that I did not make any mistakes, I would indeed need to calculate large numbers, to make sure that: \begin{align*} 162 & = 24^{\frac{-2}{5}} \times 18^{\frac{11}{5}} \\ 162^5 \times 24^2 & = 18^{11} \end{align*} These are indeed large numbers, so I'd rather avoid this, trust my prime factorisations, and instead check that: \begin{align*} (2\times3^4)^5 & = (2^3\times3)^{-2} \times (2\times3^2)^{11} \\ 2^5\times 3^{20} & = 2^{-6 + 11}\times 3^{-2 + 22} \\ \end{align*} which all looks correct.

In conclusion, for any logarithm, regardless of its base: $$\log 162 = \frac{1}{5}(-2 \log 24 + 11 \log 18)$$

Answer 2

Here's my take for solving $\log 162$ after referencing various answers and hints.

$\log 162$ can be expressed in the form of $xa + yb + z$.

$$\begin{align*} \text{i.e. } \log 162 &= xa + yb + z \\ \log 162 &= x \log 24 + y \log 28 + z \end{align*}$$

$$\begin{align*} \because 162 &= 2 \times 3^4 \\ 24 &= 2^3 \times 3 \\ 18 &= 2 \times 3^2 \end{align*}$$

$$\begin{align*} \log (2 \times 3^4) &= x \log (2^3 \times 3) + y \log (2 \times 3^2) + z \\ \log 2 + 4 \log 3 &= 3x \log 2 + x \log 3 + y \log 2 + 2y \log 3 + z \\ \log 2 + 4 \log 3 &= (3x + y)\log 2 + (x + 2y) \log 3 + z \end{align*}$$

Comparing the coefficients,

$$\begin{equation*} \begin{cases} 3x + y = 1 \\ x + 2y = 4 \\ z = 0 \end{cases} \end{equation*}$$

Solving the system of equations,

$$\begin{align*} x = -\frac{2}{5} \\ y = \frac{11}{5} \end{align*}$$

Substituting,

$$\begin{equation*} \begin{split} \log 162 &= -\frac{2}{5}a + \frac{11}{5}b \\ &= \frac{11b - 2a}{5} \end{split} \end{equation*}$$

Thanks @Stef and @lulu!