Question

solve for n.
(log equation below.)

Answer 1

\[\log_{a} 4n-2\log_{a} x = \log_{a}x \]

Answer 2

\[\log_{a}4n = \log_{a}x + \log_{a}x ^{2} \]
now
\[\log_{a}4n = \log_{a}x ^{3} \]

Answer 3

proceed further..

Answer 4

can you explain what you did there? i dont understand

Answer 5

and also m*log a = log(a)^m

Answer 6

\[m \log_{a}x = \log_{a}x ^{m} \]

Answer 7

\[\log_{a}x + \log_{a}y = \log_{a}xy \]

Answer 8

right. i understand where the logax^2 came from. but how did you make the equation into that? because in the original question there is subtraction not addition.

Answer 9

its
\[2\log_{a}x = \log_{a}x \]

Answer 10

yes the 2 became the exponent of the x.

Answer 11

yeah......
so low its like
\[\log_{a}4n - \log_{a}x ^{2} = \log_{a}x \]

Answer 12

ohhh.

Answer 13

u can proceed now???? @NY,NY

Answer 14

lets see.
4n = x^3

Answer 15

yeah.....:)))))

Answer 16

n = (x^3)/4

Answer 17

will that be the final answer?

Answer 18

is there any constant value for "x" given??? if no then ths ur answer....

Answer 19

a value for x? no, the directions are to only solve for n.

Answer 20

its ur answer then...

Answer 21

Thank you!

Answer 22

welcome