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

Question

1918 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-23T09:45:23

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

$27$ $27$ can be written as ${(3)}^{3}$ $(3)^3$

the expression now becomes:
${(3)}^{3 . (x + 2)} = {(3)}^{1}$ $(3)^(3.(x+2)) = (3)^1$

equating exponents:
$3 . (x + 2) = 1$ $3.(x+2) = 1$

$3 x + 6 = 1$ $3x+6 = 1$

$x = - \frac{5}{3}$ $x= -5/3$

How do you solve 27^(x+2)=327x+2=327^(x+2)=3?