Question

Solve for x where x is a real number in the equation below. (show work please)

Answer 1

\[27^{2x}=9^{x-3}\]

Answer 2

Note:
\[27 = 3^3\]
\[9 = 3^2\]
so \[\large 27^{2x} = 9^{x-3}\]
\[\large \implies (3^3)^{2x} = (3^2)^{x-3}\]
\[\large \implies 3^{6x} = 3^{2(x-3)}\]
\[\large \implies 6x = 2(x-3)\]
\[\large \implies 6x = 2x - 6\]
\[\large \implies 6x - 2x = -6\]
\[\large 4x = -6\]
\[\large \implies x = -\frac 64\]
\[\large \implies x = -\frac 32\]
does that help?

Answer 3

Yes it did. Im just horrible with rules of exponents. Thank you for the help.