# solving equations with logs

• Apr 26th 2010, 09:56 AM
wahhdoe
solving equations with logs
Hi, I have a question with 2 parts regarding logs, i have done the first part and believe very close to the second part but cannot quite get there and was wondering if someone could please help me?

a) show log4(3)=log2(root3)

but i am stuck with this part:

b) solve these simultaneous equations
2log2(y)=log4(3)+log2(x) and 3^y=9^x

any help with this would be great

• Apr 26th 2010, 10:37 AM
Failure
Quote:

Originally Posted by wahhdoe
Hi, I have a question with 2 parts regarding logs, i have done the first part and believe very close to the second part but cannot quite get there and was wondering if someone could please help me?

a) show log4(3)=log2(root3)

but i am stuck with this part:

b) solve these simultaneous equations
2log2(y)=log4(3)+log2(x) and 3^y=9^x

any help with this would be great

From the second equation, $3^y=9^x$, you get $3^y=3^{2x}$, and, therefore, $y=2x$. Now plug this into the first equation. - What do you get?
• Apr 28th 2010, 02:35 PM
wahhdoe
yeah i got y=2x as well but then i get this:

log2(4x^2)-log2(x)-log2(root3)=0

and i still cannot solve this
• Apr 28th 2010, 02:42 PM
pickslides
Quote:

Originally Posted by wahhdoe
yeah i got y=2x as well but then i get this:

log2(4x^2)-log2(x)-log2(root3)=0

and i still cannot solve this

Work on the LHS with using $\log_ab-\log_ac = \log_a\frac{b}{c}$

Then change the RHS into $\log_21$