simplification of some logs, give me the fastest approach:

log(10/10^m)

and log(1/root4over10)

Question

simplification of some logs, give me the fastest approach:

log(10/10^m)

and log(1/root4over10)

anonymous · Answer

|dw:1413102966117:dw| yeah this is the next step right? is it correct?

anonymous · Answer

But what do I do after?... it's just 1(10^1-m)

alekos · Answer

do you still need help?

anonymous · Answer

yes please sir

anonymous · Answer

@Auxuris  :o

anonymous · Answer

oh oops

do the same thing

bring the 1-m to the left side of the log10 and then turn log10 into 1

anonymous · Answer

so it's supposed to be 1-m?

anonymous · Answer

but I still don't get how you move 1-m, I know it's that certain rule- but it just doesn't click.

anonymous · Answer

whenever your log or lg or ln has a power, you can bring it over to the other side

anonymous · Answer

but wouldn't it be 1-m(1)(10)?

alekos · Answer

it becomes  (1-m)Log10

anonymous · Answer

what happened to the 10? from 10/10^m

alekos · Answer

10/10^m becomes  10^(1-m)

alekos · Answer

do you follow?

anonymous · Answer

log10(10^1-m)...
so it's ... (1-m)(log10) but where did the 10 go?

alekos · Answer

Log(10/10^m) = Log(10^(1-m))  because 10/10^m  is the same as 10^1/10^m which equals 10^(1-m)  by the laws of exponents

alekos · Answer

can you see that now?

anonymous · Answer

not really.. ._. sorry, could you draw it? with arrows or something?

alekos · Answer

what don't you understand?

anonymous · Answer

where did the 10 go, because you need to multiply it out right?

alekos · Answer

no you don"t.

10/10^m = 10^(1-m)

happy with that?

anonymous · Answer

then it's just log10(10^1-m)

so it's 1(10^1-m)