Review question

# Can we solve the equation $\log_2 x + \log_3 x = 1 + \log_4 x$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8260

## Solution

1. If $a$, $b$, $c$ are positive and are the $p$th, $q$th, $r$th terms of a geometric progression prove that $\begin{equation*} (q-r) \log a + (r-p) \log b + (p-q) \log c = 0. \end{equation*}$
If the geometric progression has first term $A$ and common ratio $R$, then \begin{align*} a &= A R^{p-1},\\ b &= A R^{q-1},\\ c &= A R^{r-1}, \end{align*} where $A \ne 0$. By taking logarithms (to whichever base we like), we have \begin{align*} \log a &= \log A + (p-1) \log R,\\ \log b &= \log A + (q-1) \log R,\\ \log c &= \log A + (r-1) \log R. \end{align*} Consequently, we can just do lots of algebra to show that the result is true: \begin{align*} &(q-r) \log a + (r-p) \log b + (p-q) \log c \\ &\quad= \bigl( (q-r) + (r-p) + (p-q) \bigr) \log A + \bigl( (q-r)(p-1) + (r-p)(q-1) + (p-q)(r-1) \bigr) \log R \\ &\quad= ( qp - rp -q+r +rq - pq -r+p+ pr - qr -p+q) \log R \\ &\quad= 0. \end{align*}

This argument is unenlightening; it gives no idea of why the result should be true. Another way of thinking about it is to observe that

$\dfrac{b}{a} = R^{q-p}\quad\text{and}\quad \dfrac{c}{a} = R^{r-p}.$

If we now raise the first of these equations to the power of $r-p$ and the second to the power of $q-p$, we find that $\left(\dfrac{b}{a}\right)^{r-p}=\left(\dfrac{c}{a}\right)^{q-p}.$

Now taking logs gives $(r-p)(\log b-\log a)=(q-p)(\log c-\log a)$, and rearranging this gives us the required identity.

1. Solve the equation $\log_2 x + \log_3 x = 1 + \log_4 x$, giving your answer to three significant figures.

Let’s rewrite the logarithms, so that they’re all to the same base. We use the change of base formula to do this. (There is more on this topic in Changing bases.) We may as well use base $10$ for our logarithms, so we have $\log_a x=\frac{\log_{10} x}{\log_{10} a}.$

Our equation then becomes $\frac{\log_{10} x}{\log_{10} 2} + \frac{\log_{10} x}{\log_{10} 3} = 1 + \frac{\log_{10} x}{\log_{10} 4}.$ This is now a linear equation in $\log_{10} x$, so we can simply rearrange it to solve it: $\left( \frac{1}{\log_{10} 2} + \frac{1}{\log_{10} 3} - \frac{1}{\log_{10} 4} \right)\log_{10} x = 1$ so that $\log_{10} x = 1\biggm/\left(\frac{1}{\log_{10} 2} + \frac{1}{\log_{10} 3} - \frac{1}{\log_{10} 4} \right)\approx 0.266\,179$ Exponentiating both sides with base $10$ gives $x\approx 10^{0.266\,179}\approx 1.85$ to 3 significant figures.

Alternatively, we can say, let $\log_2x = a$, $\log_3x=b$, $\log_4x=c$, so the equation becomes, $a + b = 1 + c$.

Now $\log_4x = c$ implies $4^c=x = 2^a$, and as $4^c = 2^{2c}$, we have $a = 2c$. Thus the equation simplifies to $b+c=1$.

We have $x=3^b = 4^c$ and $b=1-c$, so $3^{1-c}= 4^c$. Taking logarithms of both sides (to a base of our choice) gives $(1-c)\log 3=c\log 4$, and so $c = \dfrac{\log 3}{\log 4 + \log 3}$. (A particularly nice base to pick is $3$, in which case this simplifies to $\dfrac{1}{\log_3 4+1}$.)

Thus $x = 4^c = 4^{\log 3/(\log 4 + \log 3)} = 1.85$ to $3$ significant figures.