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If [tex]$2 \log_4 p + \log_4 q^2 = 0$[/tex], then [tex]p[/tex] is given by

Sagot :

GinnyAnsweridn
Let's solve the given equation step-by-step:

The given equation is:
[tex]\[ 2 \log_4(p) + \log_4(q^2) = 0 \][/tex]

First, recall the properties of logarithms. We know that:
[tex]\[ \log_b(a^c) = c \log_b(a) \][/tex]
So, we can rewrite [tex]\(\log_4(q^2)\)[/tex] as:
[tex]\[ \log_4(q^2) = 2 \log_4(q) \][/tex]

Substitute this into the original equation:
[tex]\[ 2 \log_4(p) + 2 \log_4(q) = 0 \][/tex]

Next, we can factor out the common factor of 2:
[tex]\[ 2 (\log_4(p) + \log_4(q)) = 0 \][/tex]

Since 2 is not zero, we can divide both sides of the equation by 2:
[tex]\[ \log_4(p) + \log_4(q) = 0 \][/tex]

Now, using the property of logarithms that states:
[tex]\[ \log_b(a) + \log_b(c) = \log_b(ac) \][/tex]
we can combine the logarithms:
[tex]\[ \log_4(pq) = 0 \][/tex]

Recall that [tex]\(\log_b(a) = 0\)[/tex] implies [tex]\(a = 1\)[/tex]. Therefore:
[tex]\[ pq = 1 \][/tex]

Thus, solving for [tex]\(p\)[/tex], we get:
[tex]\[ p = \frac{1}{q} \][/tex]

So, the value of [tex]\( p \)[/tex] in terms of [tex]\( q \)[/tex] is:
[tex]\[ p = \frac{1}{q} \][/tex]
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