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
2 log(x) = log(8)log(x2) = log(8)x2 = 8x = sqrt(8) = 2.82843 (rounded)Note that only the positive square root of 8 can serve as a solution to thegiven equation, since there's no such thing as the log of a negative number.
Log(2) 8 = log(2)2^(3) = 3log(2)2 = 3 x 1 = 3 NB Any number which has the same value as the log base equals 1. Hence log)a) a = 1 log(2)2 = 1 Hence log(2)8 = 3 Antilog 8 = 2^(3) Note the transposition of the numbers.
You use the identities: log(ab) = b log(a), and log(ab) = log a + log b. In this case, 5 log42 + 7 log4x + 4 log4y = log432 + log4x7 + log4y4 = log4 (32x7y4).
No. log 20 is a positive number , so it you subtract it from log 5 you get less than log 5. However, log10 5 = 1 - log102 = 2- log1020 . or log 5 - log 20 = log 5 - log 4*5 = log 5 - (log 5 + log 4) = log 5 - log 5 - log 4 = - log 4 But we do not need to do all of these computations, because log 5 is different from log 5 - log 20 by the law of the equality that says two equals remain equal if and only if we subtract (in our case) the same thing from them.
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
G(x) = log(2x) + 2, obviously!
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
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
2
It cannot be done because the base for the second log is not given.
To express the equation ( \log E = \log m + 2 \log V ) without logarithms, we can rewrite it using exponentiation. First, we can rewrite ( 2 \log V ) as ( \log V^2 ), leading to ( \log E = \log(m \cdot V^2) ). Exponentiating both sides gives us ( E = m \cdot V^2 ).
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Original Statement:x - 1 + 2 + log(x) = 3Simplify:x + 1 + log(x) = 3Subtract 1:x + log(x) = 2Lambert W-Function:x = (W(100*ln(10))/(ln(10)) = 1.7555794993... (rounded up).This considered log(x) to be base 10 log (x).
2 log(x) + 3 log(x) = 105 log(x) = 10log(x) = 10/5 = 210log(x) = (10)2x = 100
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)
2 log(x) = log(8)log(x2) = log(8)x2 = 8x = sqrt(8) = 2.82843 (rounded)Note that only the positive square root of 8 can serve as a solution to thegiven equation, since there's no such thing as the log of a negative number.