Certainly! Let's solve the equation [tex]\(25^n + 25^3 = 25\)[/tex] step-by-step.
First, let's rewrite the equation for clarity:
[tex]\[ 25^n + 25^3 = 25 \][/tex]
We know that [tex]\(25\)[/tex] can be written as [tex]\(5^2\)[/tex]. So we can rewrite [tex]\(25\)[/tex] as [tex]\( (5^2) \)[/tex]:
[tex]\[ (5^2)^n + (5^2)^3 = 5^2 \][/tex]
Now, applying the rule of exponents [tex]\((a^m)^n = a^{mn}\)[/tex]:
[tex]\[ 5^{2n} + 5^6 = 5^2 \][/tex]
Next, let's isolate [tex]\(5^{2n}\)[/tex]. To do this, subtract [tex]\(5^6\)[/tex] from both sides of the equation:
[tex]\[ 5^{2n} = 5^2 - 5^6 \][/tex]
Simplify the right-hand side:
[tex]\[ 5^{2n} = 5^2 - 5^6 \][/tex]
[tex]\[ 5^{2n} = 5^2 - 15625 \][/tex]
We get:
[tex]\[ 5^{2n} = -15600 \][/tex]
To solve for [tex]\(n\)[/tex], we can now take the natural logarithm on both sides:
[tex]\[ \ln\left(5^{2n}\right) = \ln\left(-15600\right) \][/tex]
We know that [tex]\(\ln(a^b) = b \ln(a)\)[/tex], so:
[tex]\[ 2n \ln(5) = \ln(-15600) \][/tex]
However, since [tex]\(\ln\)[/tex] of a negative number is not defined within the real numbers, we actually express [tex]\(-15600\)[/tex] in terms of complex numbers. Let's rewrite [tex]\(-15600\)[/tex] as [tex]\(15600 \cdot e^{i \pi}\)[/tex] (Euler's formula [tex]\(e^{i \pi} = -1\)[/tex]):
[tex]\[ \ln\left(15600 e^{i \pi}\right) = \ln(15600) + \ln(e^{i \pi}) \][/tex]
[tex]\[ 2n \ln(5) = \ln(15600) + i\pi \][/tex]
Now, we can separate the expression into its real and imaginary parts:
[tex]\[ 2n \ln(5) = \ln(15600) + i\pi \][/tex]
Divide both parts by [tex]\(2 \ln(5)\)[/tex]:
[tex]\[ n = \frac{\ln(15600) + i \pi}{2 \ln(5)} \][/tex]
and separately,
[tex]\[ n = \frac{\ln(15600) - i \pi}{2 \ln(5)} \][/tex]
Thus, we have two solutions in the form of complex numbers:
[tex]\[ n = \frac{\ln(15600) + i \pi}{2 \ln(5)} \][/tex]
and
[tex]\[ n = \frac{\ln(15600) - i \pi}{2 \ln(5)} \][/tex]
So, the solutions for the equation [tex]\(25^n + 25^3 = 25\)[/tex] are:
[tex]\[ n = \frac{\ln(15600) + i \pi}{2 \ln(5)} \][/tex]
[tex]\[ n = \frac{\ln(15600) - i \pi}{2 \ln(5)} \][/tex]
