To solve the equation [tex]\( 8^{3^{x+1}} = 2^{a^{x+1}} \)[/tex], we need to rewrite the bases to see if we can equate the exponents in a useful way.
1. Rewrite the base 8:
[tex]\[ 8 \text{ is the same as } 2^3 \][/tex]
So, [tex]\( 8^{3^{x+1}} \)[/tex] can be rewritten as:
[tex]\[
(2^3)^{3^{x+1}}
\][/tex]
2. Apply the exponent rule [tex]\((a^m)^n = a^{m \cdot n}\)[/tex]:
[tex]\[
(2^3)^{3^{x+1}} = 2^{3 \cdot 3^{x+1}}
\][/tex]
Therefore, our equation now looks like:
[tex]\[
2^{3 \cdot 3^{x+1}} = 2^{a^{x+1}}
\][/tex]
3. Equate the exponents:
Since the bases are the same (both are base 2), we can set their exponents equal to each other:
[tex]\[
3 \cdot 3^{x+1} = a^{x+1}
\][/tex]
4. Simplify the exponent equation:
We notice that we need to express [tex]\(3^{x+1}\)[/tex] in a form that matches [tex]\(a^{x+1}\)[/tex]. First, observe:
[tex]\[
3^{x+1} = 3 \cdot 3^x
\][/tex]
Simplify our exponent equation:
[tex]\[
3 \cdot (3 \cdot 3^x) = a^{x+1}
\][/tex]
[tex]\[
3^{x+1} \cdot 3 = 3^{1} \cdot 3^{x+1}
\][/tex]
[tex]\[
3^{1+1} = a^{x+1}
\][/tex]
Observe that the structure suggests:
[tex]\[
(3^1) \cdot (3^1) = a^1
\][/tex]
Simplified to:
[tex]\[
3^2 = a^1
\][/tex]
[tex]\[
9 = a
\][/tex]
This shows us [tex]\(a\)[/tex] is dependent or simplifies directly to 3.
5. Conclusion:
Therefore, we conclude:
[tex]\[
a = 3 \quad \text{and} \quad x \text{ can be any real number}
\][/tex]
So, the solution implies that [tex]\( a = 3 \)[/tex] and [tex]\( x \)[/tex] can be any real number.
