Question

which is equivalent to \\(\log_{2} n = 4\\)?
\\(\bigcirc\\) \\(\log n = \frac{\log 2}{4}\\)
\\(\bigcirc\\) \\(n = \frac{\log 2}{\log 4}\\)
\\(\bigcirc\\) \\(n = \log 4 \cdot \log 2\\)
\\(\bigcirc\\) \\(\log n = 4 \log 2\\)

Explanation:

Step1: Recall Logarithm Definition

The logarithmic equation \(\log_{a}b = c\) is equivalent to \(a^{c}=b\). For \(\log_{2}n = 4\), we have \(2^{4}=n\), so \(n = 16\). Now, use the change - of - base formula \(\log_{a}b=\frac{\log_{c}b}{\log_{c}a}\) (here we can use common logarithm with base 10, i.e., \(\log x=\log_{10}x\)).

We know that \(\log_{2}n = 4\). By the change - of - base formula, \(\log_{2}n=\frac{\log n}{\log 2}=4\).

Step2: Solve for \(\log n\)

From \(\frac{\log n}{\log 2}=4\), we can multiply both sides of the equation by \(\log 2\) to get \(\log n = 4\log 2\).

Let's check the other options:

• For the first option \(\log n=\frac{\log 2}{4}\), from \(\log_{2}n = 4\Rightarrow\frac{\log n}{\log 2}=4\Rightarrow\log n = 4\log 2

eq\frac{\log 2}{4}\), so this option is wrong.

• For the second option \(n=\frac{\log 2}{\log 4}\), since \(\log 4=\log 2^{2} = 2\log 2\), then \(\frac{\log 2}{\log 4}=\frac{\log 2}{2\log 2}=\frac{1}{2}

eq16\), so this option is wrong.

• For the third option \(n=\log 4\cdot\log 2\), \(\log 4 = 2\log 2\), so \(\log 4\cdot\log 2=2(\log 2)^{2}

eq16\), so this option is wrong.

Answer:

\(\log n = 4\log 2\) (the fourth option)