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Is y equals x differentiable at origin?
Yes, but only if the domain is the real numbers. The derivative is y = 1.
What is x if negative log x equals 21. 1?
-log x = 21.1, x > 0 log x = -21.1 x = 10-21.1 x = 1/1021.1 x = 7.94328235 x 10-22
How do you isolate x in this equation 1 equals leftbracket 1 plus x right bracket exponent 5?
with something called logarithms. So 1 = (1 + x)^5 log 1 = log ((1+x)^5) log 1 = 5 x log (1 +x) but log 1 = 0 therefore 0 = 5 x log(1+x) divide both sides by 5 and you get 0 = log (1+x) we know that log 1 = 0, therefore 1+ x = 1 and so x = 0
What is the answer to Log x plus 4 minus log6 equals 1?
log(x) + 4 - log(6) = 1 so log(x) + 4 + log(1/6) = 1 Take exponents to the base 10 and remember that 10log(x) = x: x * 104 * 1/6 = 10 x = 6/1000 or 0.006
How do you solve log x plus 2 equals log 9?
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
Is log x differentiable at 1?
yes...
The value of log log4log4x equals 1 then x equals?
the value of log (log4(log4x)))=1 then x=
What is x if log x equals -3?
If the log of x equals -3 then x = 10-3 or 0.001or 1/1000.
Is y equals x differentiable at origin?
Yes, but only if the domain is the real numbers. The derivative is y = 1.
Log x plus 7 equals 1 plus Log x-1?
logx +7=1+log(x-1) 6=log(x-1)-logx 6=log[(x-1)/x] 10^6=(x-1)/x 1,000,000x=x-1 999,999x=-1 x=-1/999,999
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)
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 x if negative log x equals 21. 1?
-log x = 21.1, x > 0 log x = -21.1 x = 10-21.1 x = 1/1021.1 x = 7.94328235 x 10-22
How do you isolate x in this equation 1 equals leftbracket 1 plus x right bracket exponent 5?
with something called logarithms. So 1 = (1 + x)^5 log 1 = log ((1+x)^5) log 1 = 5 x log (1 +x) but log 1 = 0 therefore 0 = 5 x log(1+x) divide both sides by 5 and you get 0 = log (1+x) we know that log 1 = 0, therefore 1+ x = 1 and so x = 0
What is the answer to Log x plus 4 minus log6 equals 1?
log(x) + 4 - log(6) = 1 so log(x) + 4 + log(1/6) = 1 Take exponents to the base 10 and remember that 10log(x) = x: x * 104 * 1/6 = 10 x = 6/1000 or 0.006
How do you solve log x plus 2 equals log 9?
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
When does log p X equals 0?
When X is 1, regardless of the base p.
