Find the value of:-

Question

If 2^x = 3^y = 6^-z , then find the value of

 

[1÷x +1÷y+1÷z]

Answer 1

2^x=3^y=6^(-z)

log(2^x)=log(3^y)

xlog2=ylog3

x/y=log3/log2

y/x=log2/log3

1/x=log2/log3(1/y)

log(3^y)=log(6^-z)

ylog3=-zlog6

y/-z=log6/log3

-y/z=log6/log3

1/z=log6/log3(-1/y)

1/z=-log6/log3(1/y)

(1/x)+(1/y)+(1/z)=??

substitute

log2/log3(1/y)+(1/y)-log6/log3)(1/y)

factor out (-1/y)

(-1/y)(-log2/log3)-1+log6/log3)

(-1/y)(log6/log3-log2/log3-1)

(-1/y)([log6-log2]/log3-1)

(-1/y)(log[6/2]/log3-1)

(-1/y)(log3/log3-1)

(-1/y)(1-1)

(-1/y)(0)=0

[1/x+1/y+1/z]=0

Answer 2

The answer is zero.

 

2^x = 3^y = 6^-z implies that

 

2^x=6^-z

xln(2)=-zln(6)

 

(1/x)=-[ln(2)/ln(6)](1/z)

 

Similarly

 

(1/y)=-[ln(3)/ln(6)](1/z)

 

Therefore

 

[1÷x +1÷y+1÷z]=-[ln(2)/ln(6)](1/z)-[ln(3)/ln(6)](1/z)+(1/z)

 

[1÷x +1÷y+1÷z]=(1/z)[1-ln(2)/ln(6)-ln(3)/ln(6)]

 

[1÷x +1÷y+1÷z]=(1/z)[1-(ln(2)+ln(3))/ln(6)]=(1/z)[1-ln(6))/ln(6)]=(1/z)[1-1]

 

[1÷x +1÷y+1÷z]=0