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Numerical Ability
Log and Antilog
Logx16=0.8 then what is the value of x
Read Solution (Total 5)
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- x^0.8=16
x=(2^4)^(10/8)
x=2^5
x=32 - 12 years agoHelpfull: Yes(32) No(4)
- log(x)16=0.8
(x)^(0.8)=16
(x)^(4/5)=16
(x)^(4)=(16)^(5)
(x)^(4)=(16)*(16)*(16)*(16)*(16)
(x)^(4)=(32)*(32)*(32)*(32)
(x)^(4)=(32)*(4)
x=32
- 12 years agoHelpfull: Yes(17) No(5)
- logx16=0.8
log16/logx=0.8 [log base property]
logx=log16/0.8
logx=log2^4/(5/4) [0.8=8/10=5/4]
logx=(5/4)log2^4
logx=log2^5 [power property of log m/nlogx^e=(e/n)logx^m
x=2^5
x=32 - 10 years agoHelpfull: Yes(4) No(0)
- log16_x=0.8
16=x^0.8
2^4=x^0.8
(2^4)^(1/0.8)=x
2^((4*10)/8)
2^5=32 - 10 years agoHelpfull: Yes(3) No(0)
- x^0.8=16
x^4/5=2^4
x^(4/5)^(5/4)=2^(5/4)^
x^1=2^5
x=32 - 8 years agoHelpfull: Yes(0) No(0)
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