You cannot solve log x- 2 unless (i) log x - 2 is equal to some number or (ii) x is equal to some number.
X=W*(A)^T Use logarithms. T=log(A)/log(X/W)
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
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
Not quite. The log(x/y) = log(x) - log(y) In words, this reads "The log of a quotient is the difference of the log of the numerator and the log of the denominator."
Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.
You cannot solve log x- 2 unless (i) log x - 2 is equal to some number or (ii) x is equal to some number.
10x = 4.6 therefore log 10x = log 4.6 and that gives x = log 4.6
3x = 18Take the logarithm of each side:x log(3) = log(18)Divide each side by log(3):x = log(18) / log(3) = 1.25527 / 0.47712x = 2.63093 (rounded)
X=W*(A)^T Use logarithms. T=log(A)/log(X/W)
log3(x)=4 x=3^4 x=81
log(x6) = log(x) + log(6) = 0.7782*log(x) log(x6) = 6*log(x)
y = log 2x → x = 1/2 <base of log>y So: y = log102x → x = 1/210y (common logs) y = loge2x → x = 1/2ey (natural logs)
It has no specific name. For example f(x) = sin(x)/log(x) where x not equal to 1
Log base 3 of 81 is equal to 4, because 3 ^ 4 = 81. Therefore, two times log base 3 of 81 is equal to 2 x 4 = 8.
I assume you are asking how to solve a logarithmic equation. Well let's quickly review what the log function is: for the equation log(x)=y, we are saying that 10^y=x. So once you have isolated the logarithm, take the value of the base, raise it to the nth power (when 'n' is the value that the function is equal to) and set that equal to the value inside of the log.
the log of a number, X, is equal to some value , N, and by definition 10 to the N power =X 10 to any power is always positive