Question

if logb base a =10 & log (32b) base 6a=5,find the value of a

Answer 1

$a^{2} = b$
$(6a)^{5} = 32b$
Therefore, 
$(6a)^{5} = 32 a^{10}$
$6^{5} = 2^{5} a^{5}$
$2a=6$
$a=3$

Answer 2

log (b base a)=10

=> log{ (32b) base6a}=5
=> log32b/log6a=5
=> log32b=5log6a
=> log32b=5log6a
=>log32+logb=5log6+5loga
=>5log2-5log6=5loga-logb
=> 5log2-5log2-5log3=5loga-logb
=> 5log3=logb-5loga
=> 5log3=logb-loga - 4loga
=> 5log3=log (b base a) -4loga
=> 5log3-10=-4loga
=> 10-5log3=4loga
=>(10-5log3)/4=loga
=> a=10^{(10-5log3)/4 }

here 5log3 have base 10