Q: Is log x differentiable at x equals 1?

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Yes, but only if the domain is the real numbers. The derivative is y = 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

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

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

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

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the value of log (log4(log4x)))=1 then x=

If the log of x equals -3 then x = 10-3 or 0.001or 1/1000.

Yes, but only if the domain is the real numbers. The derivative is y = 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

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) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 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

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

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

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 X is 1, regardless of the base p.