To solve the equation
[tex]\[
\frac{6^{m+2} - 6^m}{6^{m+1} - 6^m} = 7
\][/tex]
we need to simplify both the numerator and the denominator step by step. Let's start with the numerator:
### Numerator: [tex]\(6^{m+2} - 6^m\)[/tex]
1. Recognize that [tex]\(6^{m+2}\)[/tex] can be written as [tex]\(6^m \cdot 6^2\)[/tex].
2. Thus, [tex]\(6^{m+2} = (6^m \cdot 36)\)[/tex].
3. Now subtract [tex]\(6^m\)[/tex]:
[tex]\[
6^{m+2} - 6^m = 36 \cdot 6^m - 6^m
\][/tex]
4. Factor out [tex]\(6^m\)[/tex]:
[tex]\[
36 \cdot 6^m - 6^m = 6^m (36 - 1) = 6^m \cdot 35
\][/tex]
### Denominator: [tex]\(6^{m+1} - 6^m\)[/tex]
1. Recognize that [tex]\(6^{m+1}\)[/tex] can be written as [tex]\(6^m \cdot 6\)[/tex].
2. Thus, [tex]\(6^{m+1} = (6^m \cdot 6)\)[/tex].
3. Now subtract [tex]\(6^m\)[/tex]:
[tex]\[
6^{m+1} - 6^m = 6 \cdot 6^m - 6^m
\][/tex]
4. Factor out [tex]\(6^m\)[/tex]:
[tex]\[
6 \cdot 6^m - 6^m = 6^m (6 - 1) = 6^m \cdot 5
\][/tex]
### Combined Fraction
Now, substitute the simplified forms of the numerator and the denominator back into the original fraction:
[tex]\[
\frac{6^m \cdot 35}{6^m \cdot 5}
\][/tex]
You can factor out [tex]\(6^m\)[/tex] from both the numerator and the denominator:
[tex]\[
\frac{6^m \cdot 35}{6^m \cdot 5} = \frac{35}{5}
\][/tex]
Now, simplify the fraction:
[tex]\[
\frac{35}{5} = 7
\][/tex]
So, the equation simplifies to:
[tex]\[
7 = 7
\][/tex]
This simplification shows that the equation holds true for all values of [tex]\(m\)[/tex]. Therefore, we conclude that:
[tex]\[
\text{The equation holds true for all } m.
\][/tex]
