I learned all the scripts provided, yet I simply cannot find a way to simplify the following equations:
4log29
What I've thought so far: 9 may be written as 32, so maybe we could do something with the base log2, but therefore it would have to be 23, not the other way round - cannot think of any other approach.
EDIT: See How alogbx=xlogba? top answer (basically it explains how to transform alogb(x) to xlogb(a))
813log64
What I've done so far: 81/3=2⇒813log64=2log64 - but I have no idea how to simplify this term. (Maybe something with 4=22 and 6=3∗2 and do something with the 2s, but I don't know of any rule that allows you to "split" bases)
EDIT: turned out to be a typo, it should have read like this 813log24 which basically is 22=4
e12log9
No idea what to do here. (Maybe try to "bring down" the natural logarithm, but how?)
EDIT:
e12log9=912log9 log9e =912log99log9elog9e =3
In words: change base, then change base in the exponent and cancel (can you put it like that in English?) so that only 3 remains.
b4logbx
Same like above - try to bring down the logarithm with the base b.
EDIT: (In thoughts) isolate the log, so that blogbx remains ⇒ this is x ⇒ the final result should be $x^4.
e3log 8x
Same here with natural logarithm again.
EDIT: Same like the previous example, final result should be (8x)3.
To prevent any misunderstandings: I'm not asking for solutions or step-by-step-instructions, just want to know if I'm heading in the right direction and if there's any rule/approach I overlooked which can be applied here. (Sadly we didn't do much logarithm-related stuff in school and at university it's considered to be foreknowledge, so they don't explain it any more)
No comments:
Post a Comment