Question

( quad 2^{a}=3^{b}=6^{-c} )
( Rightarrow a log 2=b log 3=-c log 6=k )
( Rightarrow log 2=frac{k}{a} ; log 3=frac{k}{b}, log 6=-frac{k}{c} )
Now, ( log 2+log 3=log 6 )
( Rightarrow frac{k}{a}+frac{k}{b}=frac{-k}{c} )
( Rightarrow frac{1}{a}+frac{1}{b}+frac{1}{c}=0 )
( Rightarrow frac{a b+b c+a c}{a b c}=0 )
( Rightarrow a b+b c+a c=0 )
( begin{aligned} Rightarrow & text { obtion }(a) end{aligned} )

# 23. If 2a = 3 = 6 (a) O , then ab + bc + ca = . (b) 1 (c) 2 (d) none of

Solution