Please explain a bit about Logarithm Combinations

[Dark Knight 496]

Hi!

We've just started learning Logarithms.
So far so good, but once we learned the combinations (division/multiplication/exponents), I cannot say I understand it. I mean, I could easily memorize it - yes - but I'd like to understand the following:

Why does log(b)x^2 equal 2 * log(b)x?
Why does log(ab) equal log(a) + log(b)?
Why does log(a/b) equal log(a) - log(b)?

Is there a mathematical proof, similar to the one used to explain why x^1/2 = sqrt(x)? Thanks.

 

[tkhunny]

You might be better off learning those as properties, rather than seeking a technical proof.

In any case, logarithms ARE exponents.

log(a*b) = log(a) + log(b)

Compare to

\(\displaystyle x^{a}*x^{b} = x^{a+b}\)

Description

It turned a multiplication problem into an addition problem.

 

[stapel]

To see the "why", convert back to exponential form. For instance, let's look at the product-to-sum rule:

. . . . .log_b(xy) = s
. . . . .log_b(x) = t
. . . . .log_b(y) = u

...for some "s", "t", and "u". Then the exponential form says:

. . . . .b^s = xy
. . . . .b^t = x
. . . . .b^u = y

Then:

. . . . .xy = b^t b^u

...by substitution, and then:

. . . . .xy = b^{t + u}

...by basic exponent rules. But by our definitions, we have:

. . . . .xy = b^s

...so now:

. . . . .b^s = b^{t + u} = xy

Converting back to the log form, we then get:

. . . . .log_b(xy) = t + u = log_b(x) + log_b(y)

Follow the same sort of reasoning for the other rules.

Eliz.

 

[Dark Knight 496]

Ok, thanks, that works for the division.
But what about:

How does that equal?

 

[stapel]

Use the fact that squaring is multiplication. Then apply the multiplication rule.

Eliz.

 

[Gene]

Log₂(b²) =
Log₂(b*b) =
Log₂(b) +Log₂(b) =
2Log₂(b)

 

[galactus]

Let \(\displaystyle u=a^{r}\)

Then, let's say:

\(\displaystyle u^{x}=(a^{r})^{x}\)

By the law of exponents:

\(\displaystyle u^{x}=a^{rx}\)

By definition of \(\displaystyle log_{a}\):

\(\displaystyle log_{a}u^{x}=rx\)

Since \(\displaystyle r=log_{a}u:\)

\(\displaystyle log_{a}u^{x}=xlog_{a}u\)