how do you solve
3^(2x)-3^(x+2)+8=0
i think this needs a change of base on the 8
8 = 3^log.3. 8
in general
a^x = b^(log.b. a)(x)
move the term with the negative to the other side so they are all positive
then since you have all the numbers with base 3 you can say the bases are equal therefore the exponents must be equal
and then you can equate the exponents
i get x = 2 - log.3. 8
lets rewrite this as
$\displaystyle (3^x)^2-3^23^x+8=0$
this is quadratic in $\displaystyle 3^x$ so to simplify let $\displaystyle u=3^x$
The we get
$\displaystyle u^2-9u+8=0 \iff (u-1)(u-8)=0$ so u =1 and u=8
but $\displaystyle u=3^x$ so we get
$\displaystyle 1=3^x \iff \log_3(1)= \log_33^x \iff 0=x$
$\displaystyle 8=3^x \iff \ln(8)=\ln(3^x) \iff \ln(8)=x\ln(3) \iff \frac{\ln(8)}{\ln(3)}=x$
Good luck.