You can also write this as ln(6 times 4)
It is 1.2164
m = k/ln
you need to know natural logarithms3e to the 2x-1 power = 8(2x-1) ln e = ln (8/3)ln e = 1(2x-1) = ln(8/3) = 0.982x = 1.98x = 0.99
-6
LM = 4 in LN = ? Find LN. Round the answer to the nearest tenth.
Ln 4 + 3Ln x = 5Ln 2 Ln 4 + Ln x3= Ln 25 = Ln 32 Ln x3= Ln 32 - Ln 4 = Ln (32/4) = Ln 8= Ln 2
ln/mkl;m;/lm;lok
5 + (-4) = 1 x = -4
18
9\6
3lnx - ln2=4 lnx^3 - ln2=4 ln(x^3/2)=4 (x^3)/2=e^4 x^3=2e^4 x=[2e^4]^(1/3)
codominance
If L1=1 and L2=2, we would just get the Fibonacci sequence. Recall that the Fibonacci sequence is recursive and given by: f(0)=1, f(1)=1, and f(n)=f(n-1)+f(n-2) for integer n>1. Thus, we have f(2)=f(0)+f(1)=1+1=2. If L1=1 and L2=2 then we would have L1=f(1) and L2=f(2). Since the Lucas numbers are generated recursively just like the Fibonacci numbers, i.e. Ln=Ln-1+Ln-2 for n>2, we would have L3=L1+L2=f(1)+f(2)=f(3), L4=f(4), etc. You can use complete induction to show this for all n: As we have already said, if L1=1 and L2=2, then we have L1=f(1) and L2=f(2). We now proceed to induction. Suppose for some m greater than or equal to 2 we have Ln=f(n) for n less than or equal to m. Then for m+1 we have, by definition, Lm+1=Lm+Lm-1. By the induction hypothesis, Lm+Lm-1=f(m)+f(m-1), but this is just f(m+1) by the definition of Fibonnaci numbers, i.e. Lm+1=f(m+1). So it follows that Ln=f(n) for all n if we let L1=1 and L2=2.
ln(a) = 5.3 a = e5.3
14\2
LMN = LM + MN = 3x + (2x +2) = 5x + 2