# solve for x? help me complete the problem? :D

• Sep 12th 2010, 09:57 AM
lollilikeslife
solve for x? help me complete the problem? :D
original equation (it has a base of 4):
2log4(x) + log4(3) = log4(x) - log4(2)

my work so far:
• log4(x^2) + log4(3) = log4(x) - log4(2)
combine:
• log4(3x^2) = log4 (x/2)

help me finish? solve for x? gracias! :D (please supply your solving method and work as well, so i can understand)
• Sep 12th 2010, 10:07 AM
CaptainBlack
Quote:

Originally Posted by lollilikeslife
original equation (it has a base of 4):
2log4(x) + log4(3) = log4(x) - log4(2)

my work so far:
• log4(x^2) + log4(3) = log4(x) - log4(2)
combine:
• log4(3x^2) = log4 (x/2)

help me finish? solve for x? gracias! :D (please supply your solving method and work as well, so i can understand)

So:

\$\displaystyle 3x^2 = x/2\$

etc

CB
• Sep 12th 2010, 10:36 AM
e^(i*pi)
Quote:

Originally Posted by lollilikeslife
original equation (it has a base of 4):
2log4(x) + log4(3) = log4(x) - log4(2)

my work so far:
• log4(x^2) + log4(3) = log4(x) - log4(2)
combine:
• log4(3x^2) = log4 (x/2)

help me finish? solve for x? gracias! :D (please supply your solving method and work as well, so i can understand)

You can eliminate the logs now theyre the same base: \$\displaystyle 3x^2 = \dfrac{x}{2}\$