To find [tex]\(\log_d(abc)\)[/tex] given the condition [tex]\(a^2 = b^3 = c^5 = d^6\)[/tex], let’s follow a step-by-step approach.
1. Define the Common Value:
Since [tex]\(a^2 = b^3 = c^5 = d^6\)[/tex], let's denote the common value as [tex]\(k\)[/tex]. So we have:
[tex]\[
a^2 = k, \quad b^3 = k, \quad c^5 = k, \quad d^6 = k
\][/tex]
2. Express Each Variable in Terms of [tex]\(k\)[/tex]:
To find the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex], and [tex]\(d\)[/tex] in terms of [tex]\(k\)[/tex], we can rewrite each variable as:
[tex]\[
a = k^{\frac{1}{2}}, \quad b = k^{\frac{1}{3}}, \quad c = k^{\frac{1}{5}}, \quad d = k^{\frac{1}{6}}
\][/tex]
3. Finding [tex]\(abc\)[/tex]:
Now we need to find the product [tex]\(abc\)[/tex]:
[tex]\[
abc = \left(k^{\frac{1}{2}}\right) \cdot \left(k^{\frac{1}{3}}\right) \cdot \left(k^{\frac{1}{5}}\right)
\][/tex]
Combining the exponents we get:
[tex]\[
abc = k^{\frac{1}{2} + \frac{1}{3} + \frac{1}{5}}
\][/tex]
To simplify the exponents:
[tex]\[
\frac{1}{2} + \frac{1}{3} + \frac{1}{5} = 0.5 + 0.333333... + 0.2 = 1.0333333... \approx 1.0333333333333332
\][/tex]
So:
[tex]\[
abc = k^{1.0333333333333332}
\][/tex]
4. Expressing in Terms of [tex]\(d\)[/tex]:
Since [tex]\(d = k^{\frac{1}{6}}\)[/tex], we express [tex]\(k\)[/tex] as:
[tex]\[
k = d^6
\][/tex]
Therefore:
[tex]\[
abc = (d^6)^{1.0333333333333332} = d^{6 \cdot 1.0333333333333332}
\][/tex]
5. Simplify the Exponent:
Simplifying the exponent:
[tex]\[
6 \cdot 1.0333333333333332 = 6.199999999999999
\][/tex]
So:
[tex]\[
abc = d^{6.199999999999999}
\][/tex]
6. Finding [tex]\(\log_d(abc)\)[/tex]:
The logarithm [tex]\(\log_d(abc)\)[/tex] is simply the exponent when [tex]\(abc\)[/tex] is expressed as a power of [tex]\(d\)[/tex]:
[tex]\[
\log_d(abc) = 6.199999999999999
\][/tex]
Thus, the answer is:
[tex]\[
\log _d (abc) = 6.2
\][/tex]
Among the given options, this matches with:
Option 1) [tex]\(\frac{31}{5} = 6.2\)[/tex]
So, the correct answer is:
[tex]\[
\boxed{\frac{31}{5}}
\][/tex]
