the value of log (log4(log4x)))=1 then x=
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
log (21.4 ) = 1.4 log(2) = 1.4 (0.30103) = 0.42144 (rounded)
3^(-2x + 2) = 81? log(3^(-2x + 2)) = log(81) (-2x+2)log(3) = log(81) -2x = log(81)/log(3) - 2 x = (-1/2)(log(81)/log(3)) + 1
inverse log of 2= 1/(log{10}2)= 1/(log2)=1/0.3010299=3.3219. hence answer is 3.3219
The value of log o is penis
the value of log (log4(log4x)))=1 then x=
acording to me the value is 0 because the value of log 1 at any base is always 0.
Zero!
log AB^2 log A+log B+log2
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
log (21.4 ) = 1.4 log(2) = 1.4 (0.30103) = 0.42144 (rounded)
3^(-2x + 2) = 81? log(3^(-2x + 2)) = log(81) (-2x+2)log(3) = log(81) -2x = log(81)/log(3) - 2 x = (-1/2)(log(81)/log(3)) + 1
You can't: let suppose y the power of x to obtain such a result then xy=x/2 then xy-1=1/2 (y-1) log (x) = - log(2) (if x is a positive number) y-1 = -log(2)/log(x) y = 1 - log(2)/log(x) So log function must also being used!
1
inverse log of 2= 1/(log{10}2)= 1/(log2)=1/0.3010299=3.3219. hence answer is 3.3219
log x + 2 = log 9 log x - log 9 = -2 log (x/9) = -2 x/9 = 10^(-2) x/9 = 1/10^2 x/9 = 1/100 x= 9/100 x=.09