Asked by Rowen
log3^x=p and log18^x=q show that log6^3=q divide by p-q
Answers
Answered by
Reiny
log3^x=p ---> xlog3 = p
log18^x=q --> xlog18 = q
we want to show q/(p-q) = log6^3
p-q = xlog 3 - xlog18
= x(log3 - log18)
= x(log (3/18) = x log(1/6) = x(log1 - log6)
= x(0- log6) = -xlog6
the q/(p-q)
= xlog3/(-xlog6)
= - log3/log6 ≠ log6^3
something is not right with your question.
log18^x=q --> xlog18 = q
we want to show q/(p-q) = log6^3
p-q = xlog 3 - xlog18
= x(log3 - log18)
= x(log (3/18) = x log(1/6) = x(log1 - log6)
= x(0- log6) = -xlog6
the q/(p-q)
= xlog3/(-xlog6)
= - log3/log6 ≠ log6^3
something is not right with your question.
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