Question

Hello friends, please, I need to know how to get the answer out of this exponential equation? thank you

Answer 1

`x = 3`

Explanation:

First, we will use the properties of exponents that

• `a^(x+y) = a^xa^y`
• `(a^x)^y = a^(xy) = (a^y)^x`

`9^(x-1) - 2*3^x-27 = 0`

`=> 9^(-1)*9^x - 2*3^x - 27 = 0`

`=> 1/9*(3^2)^x - 2*3^x - 27 = 0`

`=> 1/9*(3^x)^2 - 2*3^x - 27 = 0`

With these manipulations, we can see this has the form of a quadratic equation, but with `3^x` as the variable rather than just `x`. As such, we will apply the quadratic formula:

`3^x = (-(-2)+-sqrt((-2)^2-4(1/9)(-27)))/(2(1/9))`

`=(2+-sqrt(16))/(2/9)`

`=(2+-4)/(2/9)`

`=9+-18`

Note that `3^x > 0` for all real `x`. Thus the case in which `3^x = 9-18 = -9` has no real solutions. This leaves us with the case in which `3^x = 9+18`.

`3^x = 9+18`

`=> 3^x = 27`

`=> 3^x = 3^3`

The answer is pretty clear at this point, but we can justify it properly by using logarithms. We will use the following properties of logarithms:

• `log(a^x) = xlog(a)`
• `log_a(a) = 1`

With those, we have

`log_3(3^x) = log_3(3^3)`

`=> xlog_3(3) = 3log_3(3)`

`:. x = 3`

Finally, we can check our answer:

`9^(3-1) - 2*3^3-27 = 81 - 2*27 - 27 = 0`

as desired.