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

## Suggestion

1. If $a$, $b$, $c$ are positive and are the $p$th, $q$th, $r$th terms of a geometric progression prove that $\begin{equation*} (q-r) \log a + (r-p) \log b + (p-q) \log c = 0. \end{equation*}$

Typically, we’d use $a$ for the first term of the geometric progression and $r$ for the common ratio, but that would be bad here, as $a$ and $r$ have already been used in the question…

What are the $p$th, $q$th and $r$th terms of the geometric progression, anyway?

Also note that the question does not specify the base of the logarithms, so presumably this should be true for any base.

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

Hmm, all of these logs have different bases. That seems very awkward. What can we do to simplify things?