Question

if lg5=a, lg3=b when \[\lg _{30}8\] is?

By the way, lg stands for log10

Answer 1

how i think:

\[\log _{30}8=\frac{ \lg8 }{ \lg30 }=\frac{ \lg24-\lg3}{ \lg10+\lg3 }=\frac{ \lg12+\lg2-b }{ \lg2+a+b }=\] and I am not sure what to do nextt

Answer 2

hmm..
looks u need the value of lg 2 (in term of a)

Answer 3

can u ?

Answer 4

oh wrong

Answer 5

can you show me?

Answer 6

not wrong, but just need extra important point there, is the value of log 2

Answer 7

look this :
log 2 = log (10/5) = log 10 - log 5 = 1 - a

now, apply it into ur equation above

Answer 8

btw, that will be easier if u setting log 8 = log (2^3) = 3 log 2

Answer 9

and log 30 = log (2 * 3 * 5) = log 2 + log 3 + log 5

Answer 10

well, log5 isnt the same as lg5

Answer 11

so I still didnt get it, sorry

Answer 12

log 8/log 30
= log (2^3)/log(2*3*5)
= 3 log2/(log2+log3+log5)
= 3 (1-a)/(1-a+b+a)
= .... simplify

Answer 13

got it ?

Answer 14

thanks

Answer 15

well, i will analyze;)