To solve the equation [tex]\(9^{x+2} = 240 + 9^x\)[/tex] for the real values of [tex]\(x\)[/tex], follow these steps:
1. Rewrite the equation with a common base:
Recall that [tex]\(9\)[/tex] can be expressed as [tex]\(3^2\)[/tex]. Therefore, [tex]\(9^{x+2}\)[/tex] can be rewritten using the base 3.
[tex]\[9^{x+2} = (3^2)^{x+2} = 3^{2(x+2)} = 3^{2x+4}\][/tex]
Similarly, rewrite [tex]\(9^x\)[/tex]:
[tex]\[9^x = (3^2)^x = 3^{2x}\][/tex]
2. Substitute these into the original equation:
[tex]\[3^{2x+4} = 240 + 3^{2x}\][/tex]
3. Introduce a substitution:
Let [tex]\(y = 3^{2x}\)[/tex]. Then the equation becomes:
[tex]\[3^{2x+4} = 3^{2x} \cdot 3^4 = y \cdot 81\][/tex]
Substitute [tex]\(y\)[/tex] into the equation:
[tex]\[81y = 240 + y\][/tex]
4. Simplify this new equation:
[tex]\[81y - y = 240\][/tex]
[tex]\[80y = 240\][/tex]
Solving for [tex]\(y\)[/tex]:
[tex]\[y = \frac{240}{80} = 3\][/tex]
5. Back-substitute [tex]\(y\)[/tex] to find [tex]\(x\)[/tex]:
Recall that [tex]\(y = 3^{2x}\)[/tex]. So:
[tex]\[3^{2x} = 3\][/tex]
From this equation, we see that the exponents must be equal, since the bases are the same:
[tex]\[2x = 1\][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[x = \frac{1}{2}\][/tex]
Thus, the solution to the equation [tex]\(9^{x+2} = 240 + 9^x\)[/tex] is:
[tex]\[ x = \frac{1}{2} \][/tex]
