Question

\frac{3^alpha + 6^alpha}{3^alpha} = 3

Explanation:

Step1: Simplify the left - hand side

We can split the fraction \(\frac{3^{\alpha}+6^{\alpha}}{3^{\alpha}}\) into two fractions: \(\frac{3^{\alpha}}{3^{\alpha}}+\frac{6^{\alpha}}{3^{\alpha}}\).
According to the rule of exponents \(\frac{a^{m}}{a^{n}} = a^{m - n}\), for \(\frac{3^{\alpha}}{3^{\alpha}}\), we have \(3^{\alpha-\alpha}=3^{0} = 1\) (since \(a^{0}=1,a
eq0\)).
For \(\frac{6^{\alpha}}{3^{\alpha}}\), we can rewrite \(6^{\alpha}=(2\times3)^{\alpha}\), and according to the power - of - a - product rule \((ab)^{n}=a^{n}b^{n}\), so \((2\times3)^{\alpha}=2^{\alpha}\times3^{\alpha}\). Then \(\frac{6^{\alpha}}{3^{\alpha}}=\frac{2^{\alpha}\times3^{\alpha}}{3^{\alpha}} = 2^{\alpha}\) (the \(3^{\alpha}\) terms cancel out).
So the left - hand side simplifies to \(1 + 2^{\alpha}\).
The equation becomes \(1+2^{\alpha}=3\).

Step2: Solve for \(\alpha\)

Subtract 1 from both sides of the equation \(1 + 2^{\alpha}=3\):
\(2^{\alpha}=3 - 1=2\)
Since \(2^{1}=2\), we can conclude that \(\alpha = 1\).

Answer:

\(\alpha = 1\)