Can we solve $\log_3 x + 2 \log_x 3 = -3$?

Can we solve $\log_3 x + 2 \log_x 3 = -3$?

Question

1. If $a$, $b$ and $c$ are any positive numbers not equal to unity, show that

$$\begin{equation*} \log_b a \log_c b \log_a c = 1. \end{equation*}$$

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1. Find the values of $x$ which satisfy the equation

$$\begin{equation*} \log_3 x + 2 \log_x 3 = -3. \end{equation*}$$

Suggestion

How can we rewrite the equation $\log_a b = c$ in a way that does not include logarithms?

Solution

Approach 1: Rewriting logarithms as exponentials

Let $\log_b a = x$, $\log_c b = y$, and $\log_a c = z$ so that, by definition, $b^x = a$, $c^y = b$, and $a^z = c.$

Thus $a = b^x = (c^y)^x =c^{xy} = (a^z)^{xy} = a^{xyz}$, which implies that $xyz = 1$, that is, $\log_b a \log_c b \log_a c = 1.$

Approach 2: Changing the base of logarithms

If $\log_ab=c$, then $a^c=b$. If we now take logs to base $d \ne 1$, then $\log_d(a^c) = \log_d b \implies c\log_d a = \log_d b.$

So we have $c = \log_a b = \frac{\log_db}{\log_da},$ and thus we can find logs to base $a$ if we know logs to base $d \ne 1$.

This means $\log_ab\log_bc\log_ca = \dfrac{\log_db}{\log_da}\dfrac{\log_dc}{\log_db}\dfrac{\log_da}{\log_dc}=1,$ as required.

The numbers $a$, $b$ and $c$ cannot be 1 since `logs to base 1’ makes no sense.

Can this result be generalised beyond three variables?

Approach 1: Rewriting logarithms as exponentials

Let $\log_3 x = y$ and $\log_x 3 = z$, so that $3^y = x$ and $x^z = 3$. Thus $3 = x^z = (3^y)^z = 3^{yz}$, which implies that $yz = 1$.

We also have $y + 2z = -3$. Solving these two equations simultaneously gives $(-2z-3)z = 1 \implies 2z^2 + 3z + 1 = 0 \implies (2z+1)(z+1) = 0.$

Thus $z = -\dfrac{1}{2}$ or $-1$, and hence $x = \dfrac{1}{3}$ or $\dfrac{1}{9}$.

Approach 2: Changing the base of logarithms

From our work on changing bases above, we have $\log_ab=\dfrac{\log_bb}{\log_ba} = \dfrac{1}{\log_ba}$.

Thus we have

$$\begin{equation*} \log_3 x + 2 \log_x 3 = -3 \iff \log_3 x + \frac{2}{\log_3 x} = -3. \end{equation*}$$

If we take

$u = \log_3 x \ne 0$

, this equation is equivalent to

$$\begin{align*} u + \frac{2}{u} = -3 &\iff u^2 + 3u + 2 = 0 \\ &\iff (u+2)(u+1) = 0 \\ &\iff u = -1 \quad\text{or}\quad u = -2 \\ &\iff \log_3 x = -1 \quad\text{or}\quad \log_3 x = -2 \\ &\iff x = \frac{1}{3} \quad\text{or}\quad x = \frac{1}{9}. \end{align*}$$