difference between a logarithm of a product and the product of logarithms and give examples of each?

Answer 1

Hugues is absolutely brilliant by demonstrating the algebraic difference between the logarithm of a product and the product of logarithms.

The logarithm of a product, as an example, would be

log[base b](xy)

As you can see, the product (xy) is inside the logarithm, so we're taking the logarithm of a product.

The product of logarithms, as an example, would be

[ log[base b](x) ] [ log[base c](y) ]

The logarithm of a product can be decomposed using the following formula:

log[base b](ac) = log[base b](a) + log[base b](c)

Whereas, the product of logarithms may only be algebraically manipulated in the following way:

log[base b](x) log[base c](y) = log[base c]( y^(log[base b](x)) )

Examples:

1) Logarithm of a product:

log[base 3](9x^3). By the log properties, this is equal to
log[base 3](9) + log[base 3](x^3)
2 + 3log[base 3](x)

2) Product of logarithms:

log[base 9](3) * log[base 2](3)

log[base 2](3 ^ (log[base 9](3))

log[base 2](3 ^ (1/2))

Answer 2

The logarithm of a product (of given numbers) and the product of the logarithm (of those same numbers) are NOT equal. A simple counterexample should suffice:

Consider log 100 = log (10 x 10) = log (10^2) = 2 (with logs to base 10). THAT'S the logarithm of a product of 10 times 10.

However, log 10 = 1, so (log 10)(log 10) = 1 x 1 = 1. (That's the product of two logarithms of 10.)

2 does NOT equal 1.

So the logarithm of the product 100 (i.e. 10 x 10) is NOT EQUAL to the product of the logarithms of 10 (i.e. 1 x 1).

Live long and prosper.

P.S. Hugues correctly noted that:

log(x*y)=log(x)+log(y), ......(H1) while

Log(x)*log(y)=log(x^(log(y))). ......(H2)

However, he didn't note (or rectify) the APPARENT lack of symmetry in equation (H2). In fact, as the LHS clearly indicates, the ' x ' and ' y ' MUST indeed be able to be interchanged in equation (H2), so that it may be re-written as:

Log(x)*log(y) = log(x^(log(y))) = log(y^(log(x))). ......(H2 ')

This is easily checked with logs to base 10:

log (2^[log 3]) = log (2^0.47712...) = 0.14363... , and also
log (3^[log 2]) = log (3^0.30103...) = 0.14363... .

That means that 2^[log 3] = 3^[log 2], or in general

x^[log y] = y^[log x] !

I don't know about anyone else's reaction, but I find this strange kind of symmetry quite remarkable, never having seen or realized it before. So, thank you, Hugues, for the stimulation you provided in helping me to realize that.

Answer 3

Here logarithm of a product means

log ( First value * Second value )

but by product of logarithms it is

log(First value) * log (Second value) = log (First value + Second value)

Answer 4

the difference is that they are not equal and that the operation of multiplacation is not conserved. for instance ln(6) = ln(2*3) = ln(2)+ln(3). the logarithm of a product is the sum of the logarithms. and ln(6) does not equal to ln(2)*ln(3)

Answer 5

log(x*y)=log(x)+log(y)
Log(x)*log(y)=log(x^(log(y)))