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What is the inverse of Y equals 6 plus logx?
You have, y = 6 + log x anti log of it, 10y = (106) x
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
What is x if log x equals -4?
log x = -4 => x = 10-4 = 0.0001
What is x if log x equals 3?
log(x) = 3x = 10log(x) = 103 = 1,000
Solve e raised to the power of negative x equals 6?
e-x = 6Take the natural log of both sides:ln(e-x) = ln(6)-x = ln(6)x = -ln(6)So x = -ln(6), which is about -1.792.
Log x - log 6 equals log 15?
log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90
What is the inverse of Y equals 6 plus logx?
You have, y = 6 + log x anti log of it, 10y = (106) x
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
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
What is the answer to log x6?
log(x6) = log(x) + log(6) = 0.7782*log(x) log(x6) = 6*log(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.
The value of log log4log4x equals 1 then x equals?
the value of log (log4(log4x)))=1 then x=
If y equals 10 then what is then what is y equals log x?
y = 10 y = log x (the base of the log is 10, common logarithm) 10 = log x so that, 10^10 = x 10,000,000,000 = x
What is x if log x equals -4?
log x = -4 => x = 10-4 = 0.0001
What is x if log x equals 3?
log(x) = 3x = 10log(x) = 103 = 1,000
How do you simplify log6 equals x?
That can't really be simplified. I can though be rewritten: Log 6 = x is another way of saying: 10x = 6
How do you solve sin x equals 6?
First, take the inverse sine of both sides of the equation. That gives you x = sin-1(6), which is sadly undefined...in reality, but who needs that! It can be proven that sin-1(x) = -i*log[i*x + √(1-x2)] So in this case: = -i*log[i*6 + √(1-36)] = -i*log[6*i + √(-35)] = -i*log(11.916*i) = 1.57 - 2.48*i
