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The value of log log4log4x equals 1 then x equals?
the value of log (log4(log4x)))=1 then x=
Log x plus log 2 equals log 2?
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
What is the value of log 2to the power1.4?
log (21.4 ) = 1.4 log(2) = 1.4 (0.30103) = 0.42144 (rounded)
How do you solve 3 to the power of negative 2x plus 2 equals 81?
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
What is the inverse log of 2?
inverse log of 2= 1/(log{10}2)= 1/(log2)=1/0.3010299=3.3219. hence answer is 3.3219
The value of log log4log4x equals 1 then x equals?
the value of log (log4(log4x)))=1 then x=
What is log 1 to the base 1?
acording to me the value is 0 because the value of log 1 at any base is always 0.
Value of log 1?
Zero!
What is the value of the given expressionlog AB2?
log AB^2 log A+log B+log2
Log x plus log 2 equals log 2?
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
What is the value of log 2to the power1.4?
log (21.4 ) = 1.4 log(2) = 1.4 (0.30103) = 0.42144 (rounded)
How do you solve 3 to the power of negative 2x plus 2 equals 81?
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
How do you half a number by using only powers?
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!
If 3 log x - 2 log y?
1
What is the inverse log of 2?
inverse log of 2= 1/(log{10}2)= 1/(log2)=1/0.3010299=3.3219. hence answer is 3.3219
Why is the answer for log x squared equals 2 different than 2log x equals 2?
log x2 = 2 is the same as 2 log x = 2 (from the properties of logarithms), and this is true for x = 10, because log x2 = 2 2 log x = 2 log x = 1 log10 x = 1 x = 101 x = 10 (check)
What is log base 3 of (x plus 1) log base 2 of (x-1)?
The browser which is used for posting questions is almost totally useless for mathematical questions since it blocks most symbols.I am assuming that your question is about log base 3 of (x plus 1) plus log base 2 of (x-1).{log[(x + 1)^log2} + {log[(x - 1)^log3}/log(3^log2) where all the logs are to the same base - whichever you want. The denominator can also be written as log(3^log2)This can be simplified (?) to log{[(x + 1)^log2*(x - 1)^log3}/log(3^log2).As mentioned above, the expression can be to any base and so the expression becomesin base 2: log{[(x + 1)*(x - 1)^log3}/log(3) andin base 3: log{[(x + 1)^log2*(x - 1)}/log(2)
