# Urgent help with logs needed!!!

• Apr 25th 2007, 02:07 AM
LoveDeathCab
Urgent help with logs needed!!!
Hey, the question is:

7(i) Express each of the following in terms of log_10 x and log_10 y.

(a) log_10 (x/y)

(b) log_10 (10x^2y)

(ii) Given that

2 log_10 (x/y)= 1+ log_10 (10x^2y)

find the value of y correct to 3 decimal places.

If someone could help me with this i'd be most grateful! :D
• Apr 25th 2007, 04:03 AM
CaptainBlack
Quote:

Originally Posted by LoveDeathCab
Hey, the question is:

7(i) Express each of the following in terms of log_10 x and log_10 y.

(a) log_10 (x/y)

(b) log_10 (10x^2y)

(ii) Given that

2 log_10 (x/y)= 1+ log_10 (10x^2y)

find the value of y correct to 3 decimal places.

If someone could help me with this i'd be most grateful! :D

Laws of logs:

log_c(a) = b means that c^b=a

log(a.b) = log(a) + log(b)

log(a/b) = log(a) - log(b)

log(a^b) = b log(a).

So:

(a) log_10 (x/y) = log_10(x) - log_10(y)

(b) log_10 (10x^2y) = log_10(10) + log_10(x^2 y) = log_10(10) + log_10(x^2) + log_10(y)

........................... = log_10(10) + 2 log_10(x) + log_10(y)

Now from the definition of logs log_a(a) = 1, so log_10(10)=1, so:

log_10 (10x^2y) = 1 + 2 log_10(x) + log_10(y)

(ii) Given that

2 log_10 (x/y)= 1+ log_10 (10x^2y)

using part (a) above on the left hand side, and (b) on the right hand side:

2 [log_10(x) - log_10(y)] = 1+1 + 2 log_10(x) + log_10(y) = 2 + 2 log_10(x) + log_10(y),

so:

-3 log_10(y) = 2

or: log_10(y) = -2/3, so y=10^(-2/3) or : y ~=0.215.

RonL