Question

solve for x:
27^(2x)=(1/9)^(x-3)

Answer 1

trick is to write using the same base
\[27=3^3\] and
\[\frac{1}{9}=3^{-2}\]

Answer 2

x= 3/4

Answer 3

then you have
\[(3^3)^{2x}=(3^{-3})^{x-3}\]

Answer 4

x=3/4

Answer 5

typo there solve
\[(3^3)^{2x}=(3^{-2})^{x-3}\] i.e.
\[6x=-2x+6\]

Answer 6

i was 1st

Answer 7

=3^6x=3^(-2x+6)
6x=-2x+6(after using exponent rule)
8x=6
x=3/4

Answer 8

\[(3^3)^{2x}=\frac{1}{(3^2)^{(x-3)}}\]
\[(3^2)^{3x}=\frac{1}{(3^2)^{(x-3)}}\]
\[(3^2)^{3x}*(3^2)^{(x-3)}=1\]
\[(3^2)^{3x+x-3}=(3^2)^0\]
\[4x-3=0\]
\[4x=3\]
\[x=\frac{3}{4}\]

Answer 9

Please let me know if any of the steps were unclear.