(21) 2(3ˣ)-18(3⁻ˣ)=9.
Multiply through by 3ˣ:
2(3²ˣ)-18=9(3ˣ),
2(3²ˣ)-9(3ˣ)-18=0.
It makes it look easier if we let y=3ˣ:
2y²-9y-18=0=(2y+3)(y-6).
So y=-3/2 and y=6 are the roots.
That means 3ˣ=-3/2, 3ˣ=6.
We can discard 3ˣ=-3/2 as an extraneous solution because raising a positive number to any power cannot be negative.
That leaves 3ˣ=6. Take logs (to any base):
xlog(3)=log(6), x=log(6)/log(3)=1.631 approx.
(22) log(x-6)+log(x-3)-log(x+2)=0.
Since log(1)=0, we can write:
log(x-6)+log(x-3)-log(x+2)=log(1),
log((x-6)(x-3)/(x+2))=log(1),
(x-6)(x-3)/(x+2)=1,
x²-9x+18=x+2,
x²-10x+16=0=(x-2)(x-8).
So x=2 or x=8.
But x=2 is an extraneous solution because it is not possible to take the logs of x-6 or x-3 because log(-4) and log(-1) don’t exist.
That leaves x=8 as the solution:
log(2)+log(5)-log(10)=0,
log(2×5)=log(10), which is true.