How do you solve $3^{x - 4} = 27^{x + 2}$ $3^(x-4)= 27^(x+2)$ ?

Question

3866 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-06-16T14:07:02

The answer is $x = - 5$ $x=-5$

$3^{x - 4} = 27^{x + 2}$ $3^(x-4)=27^(x+2)$

First step is to write $27^{x + 2}$ $27^(x+2)$ as the power of $3$ $3$

$3^{x - 4} = 3^{3 (x + 2)}$ $3^(x-4)=3^(3(x+2))$

Now you can write this equation as the equation of exponents:

$x - 4 = 3 (x + 2)$ $x-4=3(x+2)$

$x - 4 = 3 x + 6$ $x-4=3x+6$

$x - 3 x = 6 + 4$ $x-3x=6+4$

$- 2 x = 10$ $-2x=10$

$x = - 5$ $x=-5$

How do you solve 3^(x-4)= 27^(x+2)3x−4=27x+23^(x-4)= 27^(x+2)?