To solve the expression [tex]\(\log_9 3 + \log_{27} 3\)[/tex], we need to break it down step-by-step.
1. Understand the logarithmic definitions:
The logarithm [tex]\(\log_b a\)[/tex] is the power to which the base [tex]\(b\)[/tex] must be raised to yield [tex]\(a\)[/tex].
2. Evaluate [tex]\(\log_9 3\)[/tex]:
- We need to find [tex]\(x\)[/tex] such that [tex]\(9^x = 3\)[/tex].
- Note that [tex]\(9\)[/tex] can be rewritten as [tex]\(3^2\)[/tex].
- Thus, [tex]\(9^x = (3^2)^x = 3^{2x}\)[/tex].
- We need [tex]\(3^{2x} = 3\)[/tex].
- Therefore, [tex]\(2x = 1 \implies x = \frac{1}{2}\)[/tex].
Hence, [tex]\(\log_9 3 = \frac{1}{2}\)[/tex].
3. Evaluate [tex]\(\log_{27} 3\)[/tex]:
- We need to find [tex]\(y\)[/tex] such that [tex]\(27^y = 3\)[/tex].
- Note that [tex]\(27\)[/tex] can be rewritten as [tex]\(3^3\)[/tex].
- Thus, [tex]\(27^y = (3^3)^y = 3^{3y}\)[/tex].
- We need [tex]\(3^{3y} = 3\)[/tex].
- Therefore, [tex]\(3y = 1 \implies y = \frac{1}{3}\)[/tex].
Hence, [tex]\(\log_{27} 3 = \frac{1}{3}\)[/tex].
4. Sum the logarithms:
[tex]\[
\log_9 3 + \log_{27} 3 = \frac{1}{2} + \frac{1}{3}
\][/tex]
5. Add the fractions:
- Find a common denominator for [tex]\(\frac{1}{2}\)[/tex] and [tex]\(\frac{1}{3}\)[/tex]. The common denominator is 6.
- Rewrite the fractions: [tex]\(\frac{1}{2} = \frac{3}{6}\)[/tex] and [tex]\(\frac{1}{3} = \frac{2}{6}\)[/tex].
- Add the fractions:
[tex]\[
\frac{3}{6} + \frac{2}{6} = \frac{5}{6}
\][/tex]
Therefore:
[tex]\[
\log_9 3 + \log_{27} 3 = \frac{5}{6}
\][/tex]
To summarize, the solution to the expression [tex]\(\log_9 3 + \log_{27} 3\)[/tex] is [tex]\(\frac{5}{6}\)[/tex].
