7. EXPONENTIAL BASE 2:
23 = 2 × 2 x 2 = 8
The LOGARITHM BASE 2 OF 8
goes the OTHER WAY:
Enzo Exposyto 16
8. The logarithm base 2 of 8 is 3,
BECAUSE
2 cubed is 8;
so the logarithm base 2 of 8 is 3:
Enzo Exposyto 17
9. THE LOGARITHM BASE 2 OF 8
IS THE OPERATION THAT ALLOWS US
OF GOING BACK TO THE EXPONENT 3
EXPONENTIAL 2x and LOGARITHM BASE 2
EXPONENT x 2x
1 2
2 4
3 8
4 16
Enzo Exposyto 18
10. In other words …
THE LOGARITHM BASE 2 OF 8
IS
THE EXPONENT 3
… MORE PRECISELY …
THE LOGARITHM "3"
IS THE EXPONENT
WHICH WE HAVE TO PUT ON
THE BASE “2”
TO GET “8”
Enzo Exposyto 19
11. Now, since
log2(8) = 3 and 8 = 23
then
log2(23) = 3
We can see that
THE LOGARITHM "3"
IS THE EXPONENT
WHICH WE HAVE TO PUT ON
THE BASE “2”
TO GET “23”
Enzo Exposyto 20
12. EXPONENTIAL and LOGARITHM
with the same base
CANCEL EACH OTHER.
This is true because
exponential and logarithm
with the same base
are INVERSE OPERATIONS
It is just like
Addition and Subtraction,
Multiplication and Division,
Exponentiation and Root, …
when they're
INVERSE OPERATIONS
Enzo Exposyto 21
13. Now, we can introduce
the ANTILOGARITHM BASE b:
antilogb(x) = bx
It's, simply, an EXPONENTIAL
and represents the antilogarithm
when we operate
with a logarithm …
It’s such that
logb(antilogb(x)) = x
The meaning is
logb(bx) = x
Enzo Exposyto 22