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

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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.

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