Answer:
2
Step-by-step explanation:
This is a quadratic in terms of [tex]3^x[/tex].
[tex](3^x)^2-6(3^x)-27=0[/tex]
I'm going to substitute [tex]u=3^x[/tex]:
[tex]u^2-6u-27=0[/tex]
This is actually factorable since all you have to do is find two numbers that multiply to be -27 and add to be -6.
These numbers are -9 and 3 since (-9)(3)=-27 while -9+3=-6.
[tex](u-9)(u+3)=0[/tex]
This implies that either u-9=0 or u+3=0.
u-9=0 when u=9. (I added 9 on both sides here.)
u+3=0 when u=-3. (I subtracted 3 on both sides here.)
Recall the substitution:
[tex]u=3^x[/tex]
So replacing our solutions that are in terms of u to in terms of x:
[tex]3^x=9[/tex] or [tex]3^x=-3[/tex]
The second equation has no real solution. [tex]3^x>0[/tex] for all x.
So there is no way you find an x such that [tex]3^x[/tex] would be negative.
We only need to solve:
[tex]3^x=9[/tex]
[tex]3^x=3^2[/tex]
[tex]x=2[/tex]
Check:
Replace x with 2 in given problem:
[tex]3^{2\cdot 2)-6(3^2)-27=0[/tex]
[tex]3^4-6(9)-27=0[/tex]
[tex]81-54-27=0[/tex]
[tex]27-27=0[/tex]
[tex]0=0[/tex] which is a true equation so x=2 checks out.
