How to express this equation in terms of p and q ?

Question

Given #log_7 x^2 y = p# and #log_7 x^2 y^4 = 2q#, express #log_7 xy^(1/3)# in terms of p and q.

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Answer 1 · 2017-09-03T17:55:28

#log_7 xy^(1/3)=(5p-q)/3#

As #log_7 x^2y=p# .........(1)

and #log_7 x^2y^4=2q# .......(2)

#log_7 x^2y^4-log_7 x^2y=(2q-p)#

or #log_7 y^3=2q-p#

and #log_7 y=(2q-p)/3# .......(3)

hence #log_7 y^(1/3)=(2q-p)/9# .......(4)

From (1) and (3) #log_7 x^2=p-(2q-p)/3=(4p-2q)/3# ......(5)

and #log_7 x=(4p-2q)/3xx1/2=(2p-q)/3# ......(6)

and from (4) and (6)

#log_7 xy^(1/3)=(2p-q)/3+(2q-p)/9=(5p-q)/9.#

Answer 2 · 2017-09-03T18:10:26

#" The Reqd. Exp.="1/9(5p-q).#

Using the Familiar Rules of #log# function, we have,

#log_7 x^2y=p rArr log_7 x^2+log_7y=p..................................(1).#

#log_7x^2y^4=2q rArr log_7x^2+4log_7y=2q...........................(2).#

#:. (2)-(1) rArr 3log_7y=2q-p.#

# :. log_7y=1/3(2q-p).....................................................(3).#

Then, #(3), and, (1) rArr log_7x^2=2log_7x=p-1/3(2q-p),#

#:. log_7x=1/2{1/3(3p-2q+p)}=1/6(4p-2q),#

# :. log_7x=1/3(2p-q).....................................................(4).#

Therefore, the Reqd. Exp. =#log_7xy^(1/3),#

#=log_7x+1/3log_7y,#

#=1/3(2p-q)+1/3{1/3(2q-p)},#

#=1/9{3(2p-q)+(2q-p)},#

# rArr" the Reqd. Exp.="1/9(5p-q).#