A conceptual doubt in Logarithms?

Question

If we have log of b to the base a, do 'a' and 'b' always have to take positive values? Why is it so?

Answer 1

We start with ln which is how all other logarithms are defined.

ln(y)=x <=> e^x=y

So ln(y) =x means that e^x =y now even if x is negative y will always be positive so there is no x such that e^x=y can be negative. In fact y > 0.

log_a(b) ( log base a of b ) is defined as ln(b)/ln(a) where ln is log base e.

So to say that log_a(b) =x means that ln(b)/ln(a) = x but notice that log_a(b) is defined in terms of ln(b) and ln(a). If a and b are negative then ln(b) or ln(a) doesn't make any sense. So the base and the argument must both be positive.

Answer 2

In fact there are two different questions involved:
In log of b to the base a why always:
(1) a have to take positive values?
(2) b have to take positive values?

(Assuming the discaussion takes place in real numbers)

Let x=log of b to the base a
Then we have a^x=b.

Answer to (1):
although a^x may be defined for some negative a, but the function y=a^x fails to be a continuous real function. Just mention (-1)^(1/2). Thus avoid negative bases when you are working with logarithms.

Answer to (2):
when a is positive then b=a^x IS positive!

Answer 3

It's possible to have negative basis when a and b are integer. For example, log of -8 to base -2 = 3, because (-2)^3 = -8. But log 8 to base -2 does not exist.

II b<0 and a is not an integer, the log a to base b is not defined. In order for the log function to be well defined on R, b has to be positive.

Answer 4

a>0 and a<>1. If a<0 there would be many problems with the function, for example we can't take an even root of a negative number so any number like 1/2 would be excluded from the range.
b>0 because a positive number raised to a power is never negative, which means there is no power we could raise a>0 to to get a negative number b, hence no log b.

Answer 5

logarithms is essentially the inverse function of the exponential....

look at the graph of

y = 2^x

you will notice that "x" can be anything from -infiity to +infinity

BUT: "y" is always positive

- - - -
inverse function:
x = 2^y
log x = y log(2)
y = log(x) / log(2)

"y equals log(x) base(2)"

notice that the domain & range have swapped
"y" can range from - infinty to + infinity
"x" must be greater than zero

- - - - - -

note if you look at the graph of

y = (-2)^x it is not continuous
for example x = 1/2
y = (-2)^1/2 = sqrt (-2) ... which is a "complex number"

this can be dealt with, but for ANY "negative base", serious issues arise with respect to coninuity and "closure"

hence the "domain and range" tidiness are typically observed

Answer 6

Nope, don't follow rules too strictly, you can say that a and b are negative using exponents.

(-4)^3 = -64

log(-4)-64 = 3

So we have an example saying that we can have a negative base and a negative... the spot where -64 is.

We can also have a negative base with positive "normal script." Normal script isn't actually a thing, but I'll use it to represent the spot where -64 is because I don't know the name of that place.

(-4)^2=16

log(-4)16 = 2

So, it doesn't mean a and b have to be positive.

Answer 7

Because no power can yield a negative number.

Example:

a^(-2) = 1/a^2
a^ 0 = 1
a^2 = a^2