Respuesta :
Answer:
N=3
Step-by-step explanation:
Let's solve the given equation:
\[ \frac{5^{n+3}}{25^{2n-3}} = 5^0 \]
We know that \(5^0 = 1\), so the equation becomes:
\[ \frac{5^{n+3}}{25^{2n-3}} = 1 \]
Now, let's express \(25\) as \(5^2\):
\[ \frac{5^{n+3}}{(5^2)^{2n-3}} = 1 \]
Simplify the expression:
\[ \frac{5^{n+3}}{5^{4n-6}} = 1 \]
Apply the rule \(a^m / a^n = a^{m-n}\):
\[ 5^{n+3-(4n-6)} = 1 \]
Combine like terms:
\[ 5^{-3n+9} = 1 \]
Now, equate the exponent to zero:
\[ -3n+9 = 0 \]
Solve for \(n\):
\[ -3n = -9 \]
\[ n = 3 \]
So, \(n = 3\) satisfies the given equation.
Answer:
n=3
Step-by-step explanation:
I'm going to guess that you are saying [tex]\dfrac{5^{n+3}}{25^{2n-3}}=5^0[/tex].
First of all, [tex]a^0=1[/tex] for any [tex]a.[/tex]
[tex]\dfrac{5^{n+3}}{25^{2n-3}}=1[/tex]
Multiply both sides by [tex]\left(25^{2n-3}\right)\neq0[/tex].
[tex]5^{n+3}=25^{2n-3}\implies 5^{n+3}=\left(5^2\right)^{2n-3}[/tex]
Use the exponent rule: [tex]\left(a^m\right)^n=\left(a^n\right)^m=a^{nm}[/tex].
[tex]5^{n+3}=5^{2(2n-3)}\implies 5^{n+3}=5^{4n-6}[/tex]
Use the exponent rule: [tex]a^b=a^c\implies b=c[/tex]
[tex]n+3=4n-6\implies n-4n=-6-3\implies -3n=-9\implies \boxed{n=3}[/tex]
Checking:
[tex]25^{2n-3}\neq0\\\text{Plug in }n=3\implies 25^{2\times3-3}=25^{6-3}=25^3=15625\ne0[/tex]
Therefore, [tex]n=3.[/tex]
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