1. ## Log help

Hi ,

Could use some advise on ideas on how to solve this math question;
log14=a log15=b and log 16=c

Now I have to write an expression for each of the following in terms of a, b and c

log 2 ( solved ) = c/4
log 3 = (help me out here)
log 4= c/2
log 5= (help again)
log6= help one more time
log 7= a-(c/4)
log 8= c-(c/4)
log 9= again help......

If anyone can point me in a direction to help solve the missing expressions I sure could use it. Have been working hours on this....

Marty

2. Originally Posted by Mjay
Hi ,

Could use some advise on ideas on how to solve this math question;
log14=a log15=b and log 16=c

Now I have to write an expression for each of the following in terms of a, b and c

log 2 ( solved ) = c/4
log 3 = (help me out here)
log 4= c/2
log 5= (help again)
log6= help one more time
log 7= a-(c/4)
log 8= c-(c/4)
log 9= again help......

If anyone can point me in a direction to help solve the missing expressions I sure could use it. Have been working hours on this....

Marty
For logs 3 and 5 note that we have b = log(15) = log(3 * 5).

For log 9 note that 9 = 3^2, so log 9 = log (3^2) = 2 * log(3)...

-Dan

3. Ok,

so then I calculate that 2(b-log5)=log3 but how do I find an expression using only a, b, c for log 5?

4. If we can assume that $\displaystyle \log(x)$ is really $\displaystyle \log_{10}(x)$ then you know that $\displaystyle \log(2)+\log(5)=1$.
Otherwise, there is something missing.

5. Thank you ! Thank you! Thank You! Completely forgot that log 10=1. Knew we were missing something.....