Question Number 149621 by mathdanisur last updated on 06/Aug/21

if 2^x =3^y =7^z =((42))^(1/3) find (1/x)+(1/y)+(1/z)=?

Answered by iloveisrael last updated on 06/Aug/21

If 2^x = 3^y = 7^z = ((42))^(1/3) , then the value of (1/x) + (1/y) + (1/z) is ___ { ((2^x =(2×3×7)^(1/3) ⇒2^(x−1/3) =21^(1/3) )),((3^y =(2×3×7)^(1/3) ⇒3^(y−1/3) =14^(1/3) )),((7^z =(2×3×7)^(1/3) ⇒7^(z−1/3) =6^(1/3) )) :} { ((x−1/3=log _2 21^(1/3) ⇒x=(1/3)(1+log _2 21))),((y−1/3=log _3 14^(1/3) ⇒y=(1/3)(1+log _3 14))),((z−1/3=log _7 6^(1/3) ⇒z=(1/3)(1+log _7 6))) :} (1/x)+(1/y)+(1/z)=(3/(log _2 42))+(3/(log _3 42))+(3/(log _7 42)) =3(log _(42) (2×3×7))= 3×1=3

Commented bymathdanisur last updated on 06/Aug/21

Thankyou Ser