The expression ( \log \left( \frac{x^2 \cdot y^3}{z^4} \right) ) can be simplified using logarithmic properties. It can be rewritten as ( \log(x^2) + \log(y^3) - \log(z^4) ). Further simplifying each term gives ( 2 \log(x) + 3 \log(y) - 4 \log(z) ). Thus, the final expression is ( 2 \log(x) + 3 \log(y) - 4 \log(z) ).
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
Take logs to base '10' Hence log(10)1024 = log(10)2^(n) log(10)1024 = nlog(10)2 n = log(10)1024 / log(10)2 n = 3.0103 / 0.30103 Divide n = 10 Hence 2^(10) = 1024
The logarithm of 0.5 to the base 2, written as log₂(0.5), is equal to -1. This is because 2 raised to the power of -1 equals 0.5 (i.e., 2^(-1) = 1/2 = 0.5). Thus, log₂(0.5) = -1.
[log2 (x - 3)](log2 5) = 2log2 10 log2 (x - 3) = 2log2 10/log2 5 log2 (x - 3) = 2(log 10/log 2)/(log5/log 2) log2 (x - 3) = 2(log 10/log 5) log2 (x - 3) = 2(1/log 5) log2 (x - 3) = 2/log 5 x - 3 = 22/log x = 3 + 22/log 5
You cannot solve log x- 2 unless (i) log x - 2 is equal to some number or (ii) x is equal to some number.
Log 2 is the exponent to which 10 must be raised to equal 2, which is approximately 0.301.
You will need 14 two's multiplied together to equal 16384. the answer to this can be found by log2(16384) = 14. Since most calculators don't have log base 2, you can do this: log(16384)/log(2) = 14. You can use the 'base 10' log or natural log [ln(16384)/ln(2) = 14]
The expression ( \log \left( \frac{x^2 \cdot y^3}{z^4} \right) ) can be simplified using logarithmic properties. It can be rewritten as ( \log(x^2) + \log(y^3) - \log(z^4) ). Further simplifying each term gives ( 2 \log(x) + 3 \log(y) - 4 \log(z) ). Thus, the final expression is ( 2 \log(x) + 3 \log(y) - 4 \log(z) ).
log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1
log(2) + log(4) = log(2x)log(2 times 4) = log(2x)2 times 4 = 2 times 'x'x = 4
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.
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.
Take logs to base '10' Hence log(10)1024 = log(10)2^(n) log(10)1024 = nlog(10)2 n = log(10)1024 / log(10)2 n = 3.0103 / 0.30103 Divide n = 10 Hence 2^(10) = 1024
When the equation 2 raised to the power of log n is simplified, it equals n.
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