Let's solve the equation [tex]\(25^{x-6} \times 5^x + 5 = 0\)[/tex].
### Step-by-Step Solution:
1. Rewrite bases with a common base:
Since [tex]\(25 = 5^2\)[/tex], we can rewrite [tex]\(25^{x-6}\)[/tex] in terms of base 5. Using this, [tex]\(25^{x-6} = (5^2)^{x-6} = 5^{2(x-6)}\)[/tex].
So, the original equation [tex]\(25^{x-6} \times 5^x + 5 = 0\)[/tex] becomes:
[tex]\[
5^{2(x-6)} \times 5^x + 5 = 0.
\][/tex]
2. Simplify the exponents:
Combine the exponents of the terms with base 5 on one side:
[tex]\[
5^{2x-12} \times 5^x + 5 = 0.
\][/tex]
Using the property [tex]\(a^m \times a^n = a^{m+n}\)[/tex], we get:
[tex]\[
5^{(2x-12) + x} + 5 = 0,
\][/tex]
which simplifies to:
[tex]\[
5^{3x-12} + 5 = 0.
\][/tex]
3. Isolate the exponential term:
Move the constant term to the other side:
[tex]\[
5^{3x-12} = -5.
\][/tex]
4. Take the logarithm:
Since [tex]\(5^{3x-12}\)[/tex] equals a negative number, we need to consider the complex solution involving logarithms.
Take the natural logarithm of both sides:
[tex]\[
\ln(5^{3x-12}) = \ln(-5).
\][/tex]
Using properties of logarithms, we get:
[tex]\[
(3x-12) \ln(5) = \ln(-5).
\][/tex]
5. Solve for [tex]\(x\)[/tex]:
The natural logarithm of a negative number includes the imaginary unit [tex]\(i\)[/tex]:
[tex]\[
\ln(-5) = \ln(5) + i\pi.
\][/tex]
Thus, the equation:
[tex]\[
(3x-12) \ln(5) = \ln(5) + i\pi.
\][/tex]
Divide both sides by [tex]\(\ln(5)\)[/tex]:
[tex]\[
3x - 12 = 1 + \frac{i\pi}{\ln(5)}.
\][/tex]
6. Solve for [tex]\(x\)[/tex]:
Isolate [tex]\(x\)[/tex]:
[tex]\[
3x = 13 + \frac{i\pi}{\ln(5)},
\][/tex]
hence,
[tex]\[
x = \frac{13}{3} + \frac{i\pi}{3 \ln(5)}.
\][/tex]
This simplifies to the solution:
[tex]\[
x = \frac{13}{3} + \frac{i\pi}{3 \ln(5)}.
\][/tex]
Additionally, considering the periodic nature of complex exponential functions, we have more solutions incorporating [tex]\(n\pi\)[/tex] where [tex]\(n\)[/tex] is an odd integer. These other solutions can be written as:
[tex]\[
x = \frac{\ln(1220703125)}{3 \ln(5)} \pm \frac{i\pi}{3 \ln(5)}.
\][/tex]
Thus the complete set of solutions are:
[tex]\[
x_1 = \frac{\ln(1220703125) - i\pi}{3\ln(5)},
\][/tex]
[tex]\[
x_2 = \frac{\ln(1220703125) + i\pi}{3\ln(5)},
\][/tex]
and
[tex]\[
x_3 = \frac{13}{3} + \frac{i\pi}{\ln(5)}.
\][/tex]
These solutions include the complex logarithmic terms and reflect the nature of the initial equation involving an exponential equating to a negative number.
