Find answers to your questions faster and easier with IDNLearn.com. Ask your questions and receive accurate, in-depth answers from our knowledgeable community members.
Sagot :
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.
### 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.
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. IDNLearn.com is your go-to source for accurate answers. Thanks for stopping by, and come back for more helpful information.
