great so far
now
y^2 - 11y + 30 = 0
(y-5)(y-6) = 0
so y = 5 or y = 6
but y = 3^x
so
3^x = 5 or 3^x = 6
I will finish the first one
take ln of both sides for
3^x = 5
ln(3^x) = ln5
x(ln3) = ln5
x = ln5/ln3
now you try the second one.
3^2x - 11(3^x) + 30 = 0 ; solve for x.
I know that you can substitute 3^x with another variable like y to make it easier, so it looks like this:
y^2 - 11y + 30 = 0
...unless I did something wrong there. Help please?
Thanks!
1 answer