Question

If `216 * 3^(2x)-73*3^x-3=0` and `u=3^x`, what is the value of `u`?

Answer 1

`u=0.375`

Explanation:

If `color(blue)u=color(magenta)(3^x)`
then
`color(white)("xxx")216 * color(magenta)(3^(color(black)2x))+(-73) * color(magenta)(3^x)-3=0`
becomes
`color(white)("xxx")216color(blue)u^2-73color(blue)u-3=0`

Using the quadratic formula (and a calculator)
`color(white)("xxx")color(blue)u=(73+-sqrt((-73)^2-4 * 216 * (-3)))/(2 * 216)`

`color(white)("xxx")=0.375`
or
`color(white)("xxx")=-0.037bar(037)`

...but if `u=3^x` then `u > 0, AA x in RR`
and therefore the negative version is extraneous,
leaving only
`color(white)("xxx")color(blue)u=0.375`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

As a bonus we could find the value of `x` (approximately)

If `3^x=u`
then
`color(white)("xxx")log_3(3^x)=log_3(u)`

`color(white)("xxx")x=log_3(u)`

`color(white)("xxx")x=log_3(0.375)`

then using a calculator
`color(white)("xxx")x~~-0.892789261`

Answer 2

`x= -0.89`

Explanation:

The equation is in the form, `au^2-bu-c` where `u=3^x`.
Since it is difficult to use `x` straightaway to solve the equation, we have replaced it with ‘u’.
Once we solve for u, we will then replace u with 3^x and find the value of x.

`216u^2-73u-3=0`
`216*3=648` can be factorised as `81*8`
`216u^2+8u-81u-3=0`
`8u*(27u+1)-3*(27u+1)=0`
`(27u+1)(8u-3)=0`
`u=-(1/27) & (3/8)`

`3^x=-(1/27)=-3^-3` This solution is imaginary.

`3^x=3/8`
`xlog3=log(3/8)=log(3)-log(8)`
`x=(log(3)/log(3))-(log(8)/log(3))=1-1.89=-0.89`