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

Sep 23, 2016

$x = \frac{3}{4}$

#### 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,

${9}^{2 x + 1} = 234 \text{ } \leftarrow$ change to the common base of 3

${3}^{2 \left(2 x + 1\right)} = {3}^{5} \text{ } \leftarrow$ multiply the indices

${3}^{4 x + 2} = {3}^{5}$

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

$4 x + 2 = 5$

$4 x = 3$

$x = \frac{3}{4}$

Sep 23, 2016

$x = \frac{3}{4}$

#### Explanation:

$243 = {9}^{2 x + 1}$
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 $\left({3}^{2}\right)$
Thus the equation becomes
${3}^{5} = {\left({3}^{2}\right)}^{2 x + 1}$
$\implies {3}^{5} = {3}^{2 \times \left(2 x + 1\right)}$
$\implies {3}^{5} = {3}^{4 x + 2}$
Since bases on both sides are equal, therefore for the equation to be true, powers must be equal.
$\implies 5 = \left(4 x + 2\right)$
Solving for $x$ we get
$x = \frac{3}{4}$