log_(10)0.01=?

log_(10)0.001=………..

log_(100)0.01

An increase in intensity level of 1 dB implies an increase in intensity of (given anti log_(10)0.1=1.2589 )

An increase in intensity level of 1 dB implies an increase in density of (given anti log_(10)0.1=1.2589 )

log_(0.1)0.01=……………..

What is the value of log_(100)0.1

Simplify: root(3)(5^((1)/(log_(7)log))+(1)/(sqrt(-log_(10)(0.1))))

solve for x:2log_(10)x-log_(x)(0.01)=5

For x>1, show that: 2log_(10)x-log_(x)0.01>=4

Simplify :""^(3)sqrt(5^((1)/(log_(7)5))+(1)/(sqrt(-log_(10)(0.1))))