Exponential and Log Functions: 3 - 4/(9^x) = 4/(81^x) = 0

[Rondell]

I have knowledge of Logs and Exponential but this question just blew me away.

\(\displaystyle 2(b)\, \mbox{ Solve }\, 3\, -\, \dfrac{4}{9^x}\, -\, \dfrac{4}{81^x}\, =\, 0\)

Please help! Thanks in advance.

 

[stapel]

I have knowledge of Logs and Exponential but this question just blew me away.

\(\displaystyle 2(b)\, \mbox{ Solve }\, 3\, -\, \dfrac{4}{9^x}\, -\, \dfrac{4}{81^x}\, =\, 0\)

What are your thoughts? What have you tried? How far did you get? Where are you stuck?

For instance, you started by applying the same methods you've used in solving other exponential equations; namely, you converted exponential terms to the same bases, getting:

. . . . .$\displaystyle 3\, -\, \dfrac{4}{9^x}\, -\, \dfrac{4}{9^{2x}}\, =\, 0$

You noticed that this fit the quadratic-type form (a linear term, a term with some sort of variable, a term with the same variable thingy squared). You multiplied through to get the variable-containing terms "up top" where they're easier to deal with, giving you:

. . . . .$\displaystyle 3\,\left(9^{2x}\right)\, -\, 4\, \left(9^x\right)\, -\, 4\, =\, 0$

You restated this more explicitly in quadratic-type form:

. . . . .$\displaystyle 3\, \left(9^x\right)^2\, -\, 4\,\left(9^x\right)\, -\, 4\, =\, 0$

You factored the quadratic by the usual means:

. . . . .$\displaystyle \bigg(3\, \left(9^x\right)\, +\, 2\bigg)\, \bigg(\left(9^x\right)\, -\, 2\bigg)\, =\, 0$

And... then what?

Please be complete. Thank you!