Please help me figure out how to solve the following:
1) 4^(x+2) = 16^(2x-2)
2) log SUB 3(5x+1) = 4
3) 4^(4x+1) = 8^(3x)
Thanks in advance!
1) $\displaystyle log(4^{x+2})=log(16^{2x-2})$
$\displaystyle (x+2)log(2^{2})=(2x-2)log(2^{4})$
$\displaystyle (2x+4)log(2)=(8x-8)log(2)$
$\displaystyle 2x+4=8x-8$
$\displaystyle 12=6x$
$\displaystyle x=2$
2) $\displaystyle 3^{4}=5x+1$
$\displaystyle 80=5x$
$\displaystyle x=16$
3) $\displaystyle log(4^{4x+1})=log(8^{3x})$
$\displaystyle (4x+1)log(2^{2})=(3x)log(2^{3})$
$\displaystyle (8x+2)log(2)=(9x)log(2)$
$\displaystyle 8x+2=9x$
$\displaystyle x=2$