logarithms are the reverse form of squares. there are different ways to solve logarithms. log b 256=4 would be b4=256. You just rotate the variables, b to the left side, 256 to the right side, and the 4 goes up. To expand a logarithm, log5 (7*11) would be log5 (7)+log5 (11) In a division logarithm, log5 (7/11), you would say log5 (7)-log5 (11) Multiplication is added, and division is subtracted. Hope that helps a little:].
o.62
x is approximately 11.18
logbase5 of x =z x=5^z
can't really work out what you're asking here but log(a) - log(b) = Log (a/b) and n·log(b) = log(bⁿ)
log5(2) + log5(10) - log5(4) = log5(20/4) = log5(5) = 1
log5 20 + log5 10 - 3log5 2 = log5 [(20*10)/(2^3)] = log5 25 = 2 (log5 5) = 2
1.268293446
log5(100) = 2.861353116 to nine decimal places.
assuming that this means log5125=x, x=3.
log5 +log2 =log(5x2)=log(10)=log10(10)=1
logarithms are the reverse form of squares. there are different ways to solve logarithms. log b 256=4 would be b4=256. You just rotate the variables, b to the left side, 256 to the right side, and the 4 goes up. To expand a logarithm, log5 (7*11) would be log5 (7)+log5 (11) In a division logarithm, log5 (7/11), you would say log5 (7)-log5 (11) Multiplication is added, and division is subtracted. Hope that helps a little:].
The base of log, if unspecified, is taken to be 10 so you would be finding the value of the logarithm of 5 to the base 10.This is the value x, such that 10^x = 5.
o.62
No, it is not.
x is approximately 11.18
logbase5 of x =z x=5^z