If loga2(3a−8)−4⋅log3a=a−2, then loga(81)=....
Explanation
Let 3a=p>0 then we can transform the equation into
loga2(3a−8)−4⋅log3a=a−2
log3a⋅loga2(3a−8)−4=a−2
log3a21⋅loga2(3a−8)−4=a−2
21log3a⋅2−4loga(3a−8)=a−2
21⋅2−4log3a⋅loga(3a−8)=a−2
−log3(3a−8)=a−2
log3(3a−8)=−(a−2)
log3(3a−8)=2−a
Convert logarithmic form to exponential form
3a−8=32−a
3a−8=3a32
p−8=p9
p2−8p=9
p2−8p−9=0
(p+1)(p−9)=0
p+1=0∪p−9=0
p=−1∪p=9
Substitute the value p=9
p=9
3a=32
a=2
Therefore
loga(81)=log2(231)=log22−3=−3