How do you solve $3^{1 - 2 x} = 243$ $3^(1-2x)=243$ ?

Question

6179 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-27T12:45:03

The answer is $x = - 2$ $x=-2$

Note that $243 = 3 \cdot 3 \cdot 3 \cdot 3 \cdot 3 = 3^{5}$ $243=3*3*3*3*3=3^5$

Our equation is

$3^{1 - 2 x} = 243$ $3^(1-2x)=243$

$\frac{3}{3^{2 x}} = 3^{5}$ $3/3^(2x)=3^5$

$3^{2 x} = \frac{3}{3^{5}} = 3^{- 4}$ $3^(2x)=3/3^5=3^(-4)$

So, $2 x = - 4$ $2x=-4$

$x = - 2$ $x=-2$

How do you solve 3^(1-2x)=24331−2x=2433^(1-2x)=243?