Sure! Let's solve the equation step-by-step. We have:
[tex]\[
\frac{7^{2p+1} - 3 \times 49^p}{4 \times 49^p} = 1
\][/tex]
First, let's simplify the expressions involving powers of [tex]\(7\)[/tex] and [tex]\(49\)[/tex]. Notice that [tex]\(49\)[/tex] can be written as [tex]\(7^2\)[/tex]:
[tex]\[49^p = (7^2)^p = 7^{2p}\][/tex]
So, let's rewrite the equation using this substitution:
[tex]\[
\frac{7^{2p+1} - 3 \times 7^{2p}}{4 \times 7^{2p}} = 1
\][/tex]
Next, factor out [tex]\(7^{2p}\)[/tex] from the numerator:
[tex]\[
\frac{7^{2p}(7 - 3)}{4 \times 7^{2p}} = 1
\][/tex]
Simplify inside the parenthesis:
[tex]\[
\frac{7^{2p} \cdot 4}{4 \times 7^{2p}} = 1
\][/tex]
Now, we can cancel [tex]\(7^{2p}\)[/tex] and 4 from the numerator and the denominator:
[tex]\[
1 = 1
\][/tex]
The equation simplifies to a true statement, which indicates that it holds for any value of [tex]\( p \)[/tex].
Therefore, [tex]\( p \)[/tex] can be any real number.
