Logs- please help im so lost (1 Viewer)

[Tams]

hey guys and gals...um well i have a maths exam tomoro
and ive been studying and all and now ive come to logs
and im so stuck
call me stupid but how on earth do u do "change of base" on a calculator?
if ur sposed to do it on a calculator at all?
say u have
log (lil 9) big 243 - (i cant write it properly)
how are u sposed to work it out?
help would be most appreciated
xxoo

 

[pc_wizz]

when you have something like:

log b
a


it is the same as (ln b)/(ln a)

for ur question just put it as (ln 243)/(ln 9)

 

[sammeh]

change of base rule is:

log_a x = (log_b x)/(log_b a)

to do this on your calculator, u first have to change to either base 10 or base e, then evaluate.

gl2u ^^

 

[Tams]

thanx heaps guys...
i get it now :uhhuh:

omg u have no idea how freaked out i was !
thanx a bizzilion

xxoo

 

[Slidey]

for ur question just put it as (ln 243)/(ln 9)

It would be a lot easier to note that 243=3^5=9^(5/2). So you have log₉9^(5/2), which is (5/2)log₉9, which is 5/2*1.

So the answer is 5/2.

 

[sammeh]

hehe slide rule has what is probly the answer they would be looking for - but u asked for the chang eof base rule, so its there.

 

[pc_wizz]

It would be a lot easier to note that 243=3^5=9^(5/2). So you have log₉9^(5/2), which is (5/2)log₉9, which is 5/2*1.

So the answer is 5/2.

getting a bit complex there

she asked how to use the change of base rule on the calc ... i showed

 

[JamiL]

It would be a lot easier to note that 243=3^5=9^(5/2). So you have log₉9^(5/2), which is (5/2)log₉9, which is 5/2*1.

So the answer is 5/2.

sorry mate not many pl no that (59049)^.5= 243
and not many ppl no that 9^5= 59049 (without using a caluculator)
so sorry, not its not obvious

 

[Slidey]

sorry mate not many pl no that (59049)^.5= 243

Thanks for stating the obvious, but you misunderstood my working. There's no need to actually compute what 9^5 is. If you know your index laws, which you definitely should, then the way I suggested is the most efficient way.

and not many ppl no that 9^5= 59049 (without using a caluculator)

I didn't even know that. The fact that you think this is a necessary step indicates you misunderstood my working.

If you have 3^x, than this is also 9^(x/2), because 9^(x/2) is really (sqrt(9))^x. And that is 3^x. So when one notes that 3^5 = 243, it follows that 9^(5/2) = 243.

so sorry, not its not obvious

I never stated it was obvious. I said it was easier, assuming you notice that 243 = 3^5.

The reason I supplied my solution was to alert the thread starter that change of base is not always the only or most efficient way to solve a question - and you can bet that since the question came out so nicely, that the examiner certainly expects some students to be able solve it without change of base.

 

[silvermoon]

geez, dont get so snarky guys - its just maths!!!! personally, i think u both rock - ur both heaps smarter than me! they're BOTH good methods and im sure its a good idea 2 know both ways 2 do it. i guess we're all a little stressed round now, but geez, calm down!

 

[Shimtek]

you definately dont change the base on a calculator

If anything your trying to avoid using the calculator

logaB=logeB/logeA or so on, thats the formula in the excel book.