If `x ,y ,z` are in G.P. and `a^x=b^y=c^z` , then `(log)_b a=(log)_a c` b. `(log)_c b=(log)_a c` c. `(log)_b a=(log)_c b` d. none of these

Options

A. `log_b a = log_ac`
B. `log_c b= log_a c `
C. `log_ba= log_b `
D. none of these

Answer

C

Explanation

x,y and z aare in G.P. Hence,
`y^(2)=xz`
We have,
`a^(x)=b^(y)=c^(z)=lamda` (say)
`rArrxloga=ylogb=zlogc=loglamda`
`rArr x=(loglamda)/(loga),y=(loglamda)/(logb),z=(loglamda)/(logc)`
Putting the values of x,y and z in (1), we get
`((lohlamda)/(logb))^(2)=(loglamda)/(loga)(loglamda)/(logc)`
`rArr(logb)^(2)=logalogc`
`rArr_(b)a=log_(e)b`