If #Log_y6=x#, #Log_y3.5=z# and #6a=3.5# then #a=??#

Question

If #Log_y6=x#, #Log_y3.5=z# and #6a=3.5# then #a=??#

#(a)# #xz#
#(b)# #z/x#
#(c)# #z+x#
#(d)# #z-x#

I searched on internet about correct answer and it was #(b)# #z/x#, but I don't know why #(b)# is correct. I think that all the answers listed are incorrect.

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Answer 1 · 2018-02-15T07:42:09

I got #a=y^z/y^x=y^(z-x)#

#6a=3.5=>a=3.5/6#

We're told that

#log_y6=x; log_y3.5=z#

We can rearrange these equations to put them into exponential form:

#y^x=6; y^z=3.5#

And substitute in the values:

#a=y^z/y^x=y^(z-x)#