Demoy Levene

2022-10-16

3^(2x) - 7(3^x) - 18 = 0

nick1337

To solve the equation ${3}^{2x}-7\left({3}^{x}\right)-18=0$ for $x$, we can make a substitution to simplify the equation.
Let's make a substitution: $u={3}^{x}$.
Now, we can rewrite the equation in terms of $u$:
${u}^{2}-7u-18=0$
To solve this quadratic equation, we can factor it:
$\left(u-9\right)\left(u+2\right)=0$
Now, we can solve for $u$:
$u-9=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}u+2=0$
$u=9\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}u=-2$
Since $u={3}^{x}$, we can substitute back to find the values of $x$:
${3}^{x}=9\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{3}^{x}=-2$
For the equation ${3}^{x}=9$, we can take the logarithm base 3 of both sides:
$x={\mathrm{log}}_{3}\left(9\right)$
$x=2$
For the equation ${3}^{x}=-2$, there are no real solutions, as ${3}^{x}$ is always positive.
Therefore, the solution to the equation ${3}^{2x}-7\left({3}^{x}\right)-18=0$ is $x=2$.

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