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

Oct 22, 2015

The solution is $x = \frac{5}{4}$.

#### 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:

${3}^{2 x - 1} = {3}^{6} / \left({\left({3}^{2}\right)}^{x + 1}\right)$

Now use the rule for power of powers: ${\left({a}^{b}\right)}^{c} = {a}^{b c}$.

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

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

${3}^{2 x - 1} = {3}^{6 - \left(2 x + 2\right)}$

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

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

$4 x = 5$, and solve for $x$ finding $x = \frac{5}{4}$