Question

How do you solve `3^(2x-1) = 729/9^(x+1)`?

Answer 1

The solution is `x=5/4`.

Explanation:

First of all, note that all numbers involved are powers of three:

• `3=3^1`;
• `9=3^2`;
• `729 = 3^6`.

So, we can rewrite the equations in terms of powers of three only:

`3^{2x-1} = 3^6/((3^2)^{x+1})`

Now use the rule for power of powers: `(a^b)^c=a^{bc}`.

`3^{2x-1} = 3^6/(3^{2x+2)`

Now use the rule for dividing powers of a same base: `a^b / a^c = a^{b-c}`:

`3^{2x-1} = 3^{6-(2x+2)}`

We're finally in the form `3^a=3^b`, which is true if and only if `a=b`, so we must solve

`2x-1 = 6-2x-2`, which we easily rearrange into

`4x=5`, and solve for `x` finding `x=5/4`