Summary.....The Big Picture : Log's and Exponential Functions 
Def'n.:  
ln(x) = ò1x  1/t dt   [ x>0]
Def'n.:  
exp (x) = y Û ln(y) = x 
exp(1) º  e [so ln(e) = 1].
exp(x) º ex
Def'n.: 
For b >0,  
bx º e xln(b). 
Note: ln(bx) = x ln(b)
Def'n.: 
For b >0,  
log b (x) = y Ûx= by
ln(1) = 0 
ln(x) > 0 for x >1 
ln(x) < 0 for 0< x <1
exp(0) = e0=1 
ex> 1 for x > 0 
0< ex < 1 for x < 0
b0 = 1 

For b > 1: 
b > 1  for x > 0 
0< bx < 1 for x <

For 0< b < 1: 
b > 1  for x < 0 
0< bx < 1 for x > 0
log b (1) = 0 
For b > 1: 
log b (x)>0 for x >1 
log b (x)<0 for 0< x <1 
For 0< b < 1: 
log b (x)<0 for x >1 
log b (x)>0 for 0< x <1
ln(A*B) = ln(A) + ln(B) eAeB =eA+B bAbB = bA+B log b (A*B) = log b (A) + log b (B)
ln(A/B) = ln(A) - ln(B) eA / eB = eA-B bA /bB = bA-B log b (A/B) = log b (A ) - log b (B)
ln(Ap/q) =p/q ln(A) (ex) p/q = e(p/q)*x (bx) p/q = b(p/q)*x log b (Ap/q )= p/q log b (A)
ln'(x) = D ln(x) = 1/
[So ln is continuous and increasing for x >0]
exp'(x) = D(ex) = ex D(bx) = ln(b) bx log b'(x) = D log b (x) = 1/( x ln(b))
  ò 1/u du = ln|u| + C   òeu du = eu  + C  ò bu du = bu / ln(b) + C  Not relevant!
As x ® ¥, ln (x) ® ¥. As x ® ¥,  ex® ¥. b >0: As x ® ¥,  bx® ¥. 

b <0:As x ®  ¥ , bx® 0
For b > 1:As x ® ¥, log b (x) ® ¥. 
For 0< b < 1:As x ® ¥, log b (x)® - ¥.
As x ® 0+ , ln(x) ® - ¥ As x ® - ¥ , ex® 0 b>0: As x ® - ¥ , bx® 0 

b< 0:As x ® - ¥,  bx® ¥.
For b > 1:As x ® 0+, ln (x) ® - ¥ 
For 0< b < 1:As x ® 0+, ln (x) ®  ¥