Solve for [tex]x[/tex]:
[tex]\[3x = 6x - 2\][/tex]

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[tex]\[\log _3 \frac{1}{\sqrt{27}} =\][/tex]
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Response:
[tex]\[\log _3 \frac{1}{\sqrt{27}} =\][/tex]



Answer :

To solve [tex]\(\log_3 \frac{1}{\sqrt{27}}\)[/tex], we can follow several steps using the properties of logarithms and exponents.

1. Simplifying the Expression Inside the Logarithm:

First, note that [tex]\(27\)[/tex] can be written as a power of 3:
[tex]\[ 27 = 3^3 \][/tex]

Next, let's rewrite [tex]\(\frac{1}{\sqrt{27}}\)[/tex] in terms of powers of 3. The square root of 27 is:
[tex]\[ \sqrt{27} = \sqrt{3^3} = 3^{3/2} \][/tex]

Thus, [tex]\(\frac{1}{\sqrt{27}}\)[/tex] can be expressed as:
[tex]\[ \frac{1}{\sqrt{27}} = \frac{1}{3^{3/2}} = 3^{-3/2} \][/tex]

2. Applying the Logarithm Properties:

Now we need to find [tex]\(\log_3 (3^{-3/2})\)[/tex]. To do this, we use the property of logarithms that states:
[tex]\[ \log_b (a^c) = c \cdot \log_b (a) \][/tex]

Here, [tex]\( a = 3 \)[/tex], [tex]\( c = -3/2 \)[/tex], and the base [tex]\( b = 3 \)[/tex]. Applying this property:
[tex]\[ \log_3 (3^{-3/2}) = -\frac{3}{2} \cdot \log_3 (3) \][/tex]

3. Evaluating Basic Logarithm:

We know that [tex]\(\log_3 (3) = 1\)[/tex] because the logarithm of a number with its own base is always 1.

4. Final Calculation:

Substituting [tex]\(\log_3 (3) = 1\)[/tex] into our equation:
[tex]\[ \log_3 (3^{-3/2}) = -\frac{3}{2} \cdot 1 = -\frac{3}{2} \][/tex]

Thus, the value of [tex]\(\log_3 \frac{1}{\sqrt{27}}\)[/tex] is [tex]\(-1.5\)[/tex].