If log b base a =10 and log 32 b base 6a = 5 then calculate the value of a
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Answered by
1
Answer:
a^{2} = b
(6a)^{5} = 32b
Therefore,
(6a)^{5} = 32 a^{10}
6^{5} = 2^{5} a^{5}
2a=6
a=3
Answered by
0
answer:
10
answer with explanation:
log (b base a)=10
=logf (32b) base6aj=5
log32b/log6a=5
>log32b=5log6a
=> log32b=5log6a
->log32logb=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]
Now, 5log3 have base 10
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