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Mathematics 18 Online
OpenStudy (anonymous):

a^b=c,b^c=a and c^a=b ,then prove that abc=a....How to solve this ??

OpenStudy (anonymous):

\[a^b=c,b^c=a , c^a=b , prove ,that , abc=a\] how to solve?

OpenStudy (anonymous):

\[a^b=c,b^c=a ,c^a=b\] obvıously we see that\[abc=a^b b^c c^a\]

OpenStudy (anonymous):

we need to fine abc =a .. How?

OpenStudy (helder_edwin):

two questions: what a, b and c are? and did u write everything correctly? (just to check)

OpenStudy (anonymous):

especially check if question says abc=a or somethıng and why dıd you rıght....

OpenStudy (anonymous):

wrıte

OpenStudy (anonymous):

if \[a ^{b }=c,b ^{c }=a,c ^{a }=b,\] then prove that abc=a

OpenStudy (anonymous):

does the questıon say wether a,b,c are posıtıve ıntergers or non posıtıve or just constants

terenzreignz (terenzreignz):

Real numbers, though.... a,b, and c ARE real numbers, right?

sam (.sam.):

\[a ^{b }=c,b ^{c }=a,c ^{a }=b \\ \\ from \\ \\ b^c=a----> b=a^{\frac{1}{c}} \\ \\ sub ~into ~ equation ~3, \\ \\ \huge c=a^{\frac{1}{ac}} \\ \\ sub ~into ~ equation ~1, \\ \\ \huge a^b=a^{\frac{1}{ac}} \\ \\ \huge a^{abc}=a \\ \\ \huge abc=1\] Am I missing something? :|

OpenStudy (anonymous):

No :)

OpenStudy (mertsj):

I also got that abc = 1 but how does that prove that a = abc?

sam (.sam.):

yay :D

terenzreignz (terenzreignz):

hey, @.Sam. \[\huge a^{abc}=a\]does not necessarily mean abc = 1 it could be that a=1

terenzreignz (terenzreignz):

On second thought, scratch that :D

OpenStudy (mertsj):

I think abc does equal 1 though. If you put each equation into log form and substitute, it comes out that abc = 1 also

terenzreignz (terenzreignz):

The way I (just) saw it was \[\large a^{abc} = a\]would imply either abc = 1 or a = 1 BUT a = 1 would imply c = 1 and b = 1 implying abc = 1 anyway so... yeah, scratch the last one :D

terenzreignz (terenzreignz):

But if a = 1, the proof is basically done (abc = a) So... abc = 1... is tricky :)

sam (.sam.):

So lame... :\, normally they'd asked abc=1, but this one's with some spinners

OpenStudy (mertsj):

\[a^b=c \rightarrow \log_{a}c=b \] \[b=c^a \rightarrow a^b=a ^{ac}\] But \[\log_{a}c=ca=c \iff a=1 \] But \[abc=1 ; a=1, \rightarrow abc=a\]

OpenStudy (mertsj):

Compliments of @Hoa (who is a very smart guy)

OpenStudy (anonymous):

@Mertsj thanks for your compliment, but pleaaaase, take the compliment out. It's totoooooooo shoooow offf :)

OpenStudy (mertsj):

@Hoa No. The truth is the truth.

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