Sure, let's prove the given equation step by step.
We start with the given expression:
[tex]\[ \frac{6^{x+2}-6^x}{6^{x+1}+6^x} \][/tex]
First, note that we can express [tex]\(6^{x+2}\)[/tex] and [tex]\(6^{x+1}\)[/tex] in terms of [tex]\(6^x\)[/tex]:
[tex]\[ 6^{x+2} = 6^x \cdot 6^2 \][/tex]
[tex]\[ 6^{x+1} = 6^x \cdot 6 \][/tex]
Substitute these into the given expression:
[tex]\[ \frac{6^x \cdot 6^2 - 6^x}{6^x \cdot 6 + 6^x} \][/tex]
Factor out [tex]\(6^x\)[/tex] from both the numerator and the denominator:
[tex]\[ \frac{6^x (6^2 - 1)}{6^x (6 + 1)} \][/tex]
Since [tex]\(6^x\)[/tex] is a common factor in both the numerator and the denominator, we can cancel it out:
[tex]\[ \frac{6^2 - 1}{6 + 1} \][/tex]
Next, evaluate the remaining terms:
[tex]\[ 6^2 = 36 \][/tex]
[tex]\[ 6 + 1 = 7 \][/tex]
So,
[tex]\[ \frac{36 - 1}{7} = \frac{35}{7} \][/tex]
Finally, divide 35 by 7:
[tex]\[ \frac{35}{7} = 5 \][/tex]
Thus, we have proven that:
[tex]\[ \frac{6^{x+2} - 6^x}{6^{x+1} + 6^x} = 5 \][/tex]
