Question

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

Explanation:

Step1: Recall the definition of logarithms

The logarithmic equation $\log_{a}b = c$ is equivalent to $a^{c}=b$. For $\log_{2}n = 4$, by the definition of logarithms, we can rewrite it as $2^{4}=n$, or $n = 16$. But we can also use the change - of - base formula or logarithmic properties. The change - of - base formula is $\log_{a}b=\frac{\log_{c}b}{\log_{c}a}$ (where $c>0,c
eq1$), and the power rule of logarithms is $\log_{a}(M^{k})=k\log_{a}M$.

Let's start from $\log_{2}n = 4$. If we take the common logarithm (base 10) of both sides, we have $\log(\log_{2}n)=\log(4)$? No, that's not the right way. Wait, we can use the change - of - base formula on $\log_{2}n$. By the change - of - base formula, $\log_{2}n=\frac{\log n}{\log 2}$. So from $\log_{2}n = 4$, we get $\frac{\log n}{\log 2}=4$.

Step2: Solve for $\log n$

Multiply both sides of the equation $\frac{\log n}{\log 2}=4$ by $\log 2$. We get $\log n=4\log 2$.

Let's check the other options:

• For the first option: From $\log_{2}n = 4$, we have $\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: Using the change - of - base formula, if we want to express $n$ in terms of common logarithms, $n = 2^{4}$, and $\log n=\log(2^{4}) = 4\log 2$, not $\frac{\log 2}{\log 4}$ (since $\log 4 = 2\log 2$, $\frac{\log 2}{\log 4}=\frac{1}{2}

eq4\log 2$), so this option is wrong.

• For the third option: $\log 4\cdot\log 2=2\log 2\cdot\log 2 = 2(\log 2)^{2}

eq\log n$ (since $\log n = 4\log 2$), so this option is wrong.

Answer:

$\log n = 4\log 2$ (the last option)