To solve the equation [tex]\( 8^{2n} = \frac{1}{64} \)[/tex], we will follow these steps:
1. Express the equation in terms of powers of the same base:
First, recognize that [tex]\( 8 \)[/tex] and [tex]\( 64 \)[/tex] can be expressed as powers of 2. We know:
[tex]\[
8 = 2^3 \quad \text{and} \quad 64 = 2^6
\][/tex]
Thus, the equation [tex]\( 8^{2n} = \frac{1}{64} \)[/tex] can be rewritten as:
[tex]\[
(2^3)^{2n} = \frac{1}{2^6}
\][/tex]
2. Simplify the exponents:
Using the power of a power property [tex]\((a^m)^n = a^{mn}\)[/tex], we rewrite the left side as:
[tex]\[
2^{6n} = \frac{1}{2^6}
\][/tex]
3. Rewrite the right side as a negative exponent:
Recall that [tex]\(\frac{1}{a^m} = a^{-m}\)[/tex]:
[tex]\[
2^{6n} = 2^{-6}
\][/tex]
4. Equate the exponents:
Since the bases are equal, the exponents must be equal. Thus:
[tex]\[
6n = -6
\][/tex]
5. Solve for [tex]\( n \)[/tex]:
[tex]\[
n = \frac{-6}{6} = -1
\][/tex]
However, from the initial transformation, there are multiple solutions due to the periodicity of exponents in the complex plane. Evaluating all the possible roots, we get:
[tex]\[
n = -1, \quad -1 + \frac{\pi i}{\ln(8)}, \quad -1 - \frac{\pi i}{\ln(8)}, \quad -1 + \frac{2 \pi i}{\ln(8)}, \quad -1 - \frac{2 \pi i}{\ln(8)}, \quad \ldots
\][/tex]
Therefore, the full set of solutions is:
[tex]\[
n = -1.00000000000000, \quad -1.0 - 3.0215734278848i, \quad -1.0 - 1.5107867139424i, \quad -1.0 + 1.5107867139424i, \quad -1.0 + 3.0215734278848i, \quad -1.0 + 4.53236014182719i
\][/tex]
In summary, the exact values for [tex]\( n \)[/tex] that satisfy the equation [tex]\( 8^{2n} = \frac{1}{64} \)[/tex] are [tex]\( -1.00000000000000 \)[/tex] and its complex counterparts:
[tex]\[
-1.0 - 3.0215734278848i, \quad -1.0 - 1.5107867139424i, \quad -1.0 + 1.5107867139424i, \quad -1.0 + 3.0215734278848i, \quad -1.0 + 4.53236014182719i
\][/tex]
