Question

a^b=b^c=ab then b+c always equal

Answer 1

Given, $a^b=b^c=ab$

$a^b=ab$

taking log both sides,

$loga^b=logab$

we know, $logAB=logA+logB$
$logA^B=BlogA$

so, $bloga=loga+logb$

$bloga-loga=logb$

$(b-1)loga=logb$

$loga^{(b-1)}=logb$

removing log from both sides,

$a^{(b-1)}=b$

similarly, $b^c=ab$

put value of b ,

$(a^{(b-1)})^c=aa^{(b-1)}$

$a^{cb-c}=a^{(b-1)+1}$

$bc - c = b$

b + c = bc

hence, b + c = bc

Answer 2

Given, a^b=b^c=abab=bc=ab 

a^b=abab=ab 

taking log both sides, 

loga^b=logablogab=logab 

we know, logAB=logA+logBlogAB=logA+logB 
logA^B=BlogAlogAB=BlogA 

so, bloga=loga+logbbloga=loga+logb 

bloga-loga=logbbloga−loga=logb 

(b-1)loga=logb(b−1)loga=logb 

loga^{(b-1)}=logbloga(b−1)=logb 

removing log from both sides, 

a^{(b-1)}=ba(b−1)=b 

similarly, b^c=abbc=ab 

put value of b , 

(a^{(b-1)})^c=aa^{(b-1)}(a(b−1))c=aa(b−1) 

a^{cb-c}=a^{(b-1)+1}acb−c=a(b−1)+1 

bc - c = bbc−c=b 

b + c = bc 

hence, b + c = bc