Question

6log9x + 3log27y=8
log3x+2log9y=2.
Solve the equation ;)

Answer 1

those numbers next to the variables aren't the bases right?

Answer 2

the equations are exactly as you have them right?

Answer 3

Yes. The numbers are the base.

Answer 4

so we have
\[6\log_9(x)+3\log_{27}(y)=8\]
and
\[\log_3(x)+2\log_9(y)=2\]
?

Answer 5

Yes!

Answer 6

Since all those bases are powers of 3, lets convert all the bases to 3 using this formula:
\[\log_{b^{n}}x= \frac{1}{n}\log_{b}x\]
9 is 3^2, and 27 is 3^3, so after converting we get:
\[\frac{6}{2}\log_{3}x+\frac{3}{3}\log_3y = 8\]
\[\log_3x+\frac{2}{2}\log_3y=2\]

Answer 7

now we have a system we can solve:
\[3\log_3x+\log_3y=8\]
\[\log_3x+\log_3y=2\]
Subtracting the second equation from the first gets rid of the y part:
\[2\log_3x=6 \Rightarrow \log_3x=3 \Rightarrow x = 27\]
Then, if:
\[\log_3x+\log_3y= 2 \Rightarrow 3+\log_3y = 2 \Rightarrow \log_3y = -1\]
So y = 1/3

Answer 8

When you get log3x=3, how do you find x?

Answer 9

lol joe i made my more complicated then it needed to be
but i did get what you did

Answer 10

When looking at log functions this is how they should be interpreted:
\[\log_b x= a \Leftrightarrow b^{a} = x\]
So when I see:
\[\log_3 x = 3\]
I change it to:
\[3^{3} = x \Rightarrow x = 27\]

Answer 11

@myininaya yeah i was scribbling on paper and it was getting ridiculous >.< haha

Answer 12

only read this if you want to get confused

Answer 13

Ok. Thx for your help! I have exams in 2 hrs xD & I got the answers! :D

Answer 14

i forgot to use that one propery ln3^3=3ln3
so i wrote it as ln3+ln3+ln3 then i was like oh yeah thats just 3ln3 lol

Answer 15

but still even if i realized that you did something much better

Answer 16

lol, I had started mine out the exact same way, and it just seemed too long, so i was grasping for straws trying to find a way to make the problem easier >.<