If \(\log_{2}(x), 1 + \log_{4}(x)\), and \(\log_{8}(4x)\) are consecutive terms of a geometric sequence, determine the possible values of \(x\).

Note:
A geometric sequence is a sequence of the form \(a\), \(ar\), \(ar^2\), \(ar^3\), \(\ldots\) , \(ar^{n-1}\), \(\ldots\) where the ratio of any term to the preceding term is constant \(\left( \dfrac{t_{n+1}}{t_n} = r,~ r \neq 0, ~n \geq 1, ~n \in \mathbb{Z} \right), ~a\neq 0\).

Solution

Since \(\log_{2}(x), 1 + \log_{4}(x)\), and \(\log_{8}(4x)\) are consecutive terms of a geometric sequence, then

Now, \(x \gt 0\) and \(4x \gt 0\), and thus \(x \gt 0\).

Also, \(\log_{2}{(x)} \neq 0\) and \(1+\log_{4}{(x)} \neq 0\). Therefore, \(x \neq 1\) or \(\frac{1}{4}\).

To solve this equation, we will convert \(\log_{4}(x)\) and \(\log_{8}(4x)\) to base \(2\) logarithms to work with a common base.

If \(\log_{2}(x), 1 + \log_{4}(x)\), and \(\log_{8}(4x)\) are consecutive terms of a geometric sequence, determine the possible values of \(x\).

Solution

So we must solve

If \(\log_{2}(x), 1 + \log_{4}(x)\), and \(\log_{8}(4x)\) are consecutive terms of a geometric sequence, determine the possible values of \(x\).

Solution

Let \(a=\log_{2}(x)\) and solve for \(a\).

So \(a=6,-2\).

Since \(a = \log_{2}(x)\),

If \(\log_{2}(x), 1 + \log_{4}(x)\), and \(\log_{8}(4x)\) are consecutive terms of a geometric sequence, determine the possible values of \(x\).

Solution

But only \(x = 64\) satisfies the restriction \(x \gt 0,\) \(x \neq 1\), \(\dfrac{1}{4}\).

Therefore, the three terms, \(\log_{2}(x)\), \(1 + \log_{4}{(x)}\), and \(\log_{8}(4x)\), form a geometric sequence when \(x = 64\).

The three terms of the sequence are \(6\), \(4\), \(\dfrac{8}{3}\), which have a common ratio of \(\dfrac{2}{3}\).

Note:
When \(x=\dfrac{1}{4}\), the three terms of the sequence (i.e. \(-2\), \(0\), and \(0\)) have a common ratio of \(0\).

A geometric sequence cannot have a common ratio of \(0\).

• The logarithmic function, \(y = \log_{c}(x)\), is the inverse of the exponential function, \(y = c^x\), where \(c \gt 0,~c \neq 1\).
• The statement \(\log_{c}(m) = n\) is equivalent to the statement \(m = c^n\).
• The logarithm, \(\log_{c}(x)\), is defined only when \(c \gt 0\), \(c \neq 1\) and \(x \gt 0\).
• A common logarithm is a logarithm base \(10\). When the base is not provided, such as \(\log (x)\), it is assumed to be a common logarithm.
• The following statements hold true for logarithms with \(c \gt 0\) and \(c \neq 1\):
  • \(c^{\log_{c}(n)} = n\), \(n \gt 0\)
  • \(\log_{c}(c^n) = n\)
  • \(\log_{c}(1)=0\)
  • \(\log_{c}(m) = \log_{c}(n)\) if and only if \(m = n\), \(m\) and \(n \gt 0\)
• For \(c \gt 0\), \(c \neq 1\), \(x \gt 0\), \(y \gt 0\),
  • \(\log_{c}(xy)= \log_{c}(x) + \log_{c}(y) \quad\quad\)(Product Law)
  • \(\log_{c}{\left( \frac{x}{y} \right)}= \log_{c}(x) - \log_{c}(y)\quad\;\)(Quotient Law)
  • \(\log_{c}(x^n) = n \log_{c}(x)\quad\quad\quad\quad\quad\;\,\)(Power Law)
• The formula for converting a logarithm from one base to another is \(\log_{c}(x) = \frac{\log_{a}(x)}{\log_{a}(c)}\). This formula is often used to convert to base \(10\).

All of the above properties and laws of logarithms can be used to simplify logarithmic expressions and solve exponential and logarithmic equations.