Question

How do you solve `243=9^(2x+1)`?

Answer 1

`x = 3/4`

Explanation:

It really is an advantage to know all the powers up to 1000.
You will then recognise that `243 = 3^5`

Without knowing this, you will have to resort to finding the prime factors. Another clue is that `9 = 3^2`

So this is an exponential equation with 3 as the base,

`9^(2x+1) = 234" "larr` change to the common base of 3

`3^(2(2x+1)) = 3^5" "larr` multiply the indices

`3^(4x+2) = 3^5`

The bases are equal, so the indices must be equal.

`4x+2 = 5`

`4x =3`

`x = 3/4`

Answer 2

`x=3/4`

Explanation:

`243=9^(2x+1)`
On inspection we see that LHS, `243` can be expressed in terms of powers of `3`,
`243=3^5`
Similarly on RHS, `9` can also be expressed as `(3^2)`
Thus the equation becomes
`3^5=(3^2)^(2x+1)`
`=>3^5=3^(2xx(2x+1))`
`=>3^5=3^(4x+2)`
Since bases on both sides are equal, therefore for the equation to be true, powers must be equal.
`=>5=(4x+2)`
Solving for `x` we get
`x=3/4`