Logarithm

[Probability]

I know logarithms are powers and I know that there are three laws or rules, but I still have questions I can't seem to answer?

1 / antilogarithm

if 5 = log₂ x,

then

how does 5 get to become the power on the RHS

thus

x = 2⁵

I can understand subtracting x from both sides to get x on the LHS, but fail to understand how 5 becomes a power on the RHS?

2/ Simplifying logs

3ln2 - ln4 = ln2³ - ln4

= ln8 - ln4

OK the concern

How do I know which side should be divided by which?

ln8 / ln4 or ln4 / ln8 ??

how do I know which way round is the correct way to divide by which ln?

 

[Bob Brown MSEE]

When I see an expression like "log₂ x"
I say to myself that it represents "The exponent of 2 that gives x as the answer"

So if 5 = log₂ x,

then

5 is "The exponent of 2 that gives x as the answer"

x = 2⁵

because that is what "log₂ x" means,

 

[JeffM]

I know logarithms are powers and I know that there are three laws or rules, but I still have questions I can't seem to answer?

1 / antilogarithm What does this mean?

if 5 = log₂ x, Until you car comfortable with logs, avoid errors by writing them as $\displaystyle log_2(x)\ rather\ than\ log_2x.$

The super-correct form will remind you that the log is a function.

then

how does 5 get to become the power on the RHS

thus

x = 2⁵

I can understand subtracting x from both sides to get x on the LHS,

You cannot get x on the LHS by subtracting. $\displaystyle y = log(x) \implies log(x) - x = y - x .$

but fail to understand how 5 becomes a power on the RHS?

The DEFINITION of a log is: $\displaystyle log_a(b) = c \iff b = a^c.$

2/ Simplifying logs

3ln2 - ln4 = ln2³ - ln4

= ln8 - ln4

OK the concern

How do I know which side should be divided by which?

ln8 / ln4 or ln4 / ln8 ?? Nothing is to be divided by any log so the issue is irrelevant.

Method 1

$\displaystyle 3ln(2) - ln(4) = ln\left(2^3\right)) - ln(4) = ln(8) - ln(4) = ln\left(\dfrac{8}{4}\right) = ln(2).$

Method 2

$\displaystyle 3ln(2) - kn(4) = 3ln(2) - ln\left(2^2\right) = 3ln(2) - 2ln(2) = ln(2) * (3 - 2) = ln(2) * 1 = ln(2).$

how do I know which way round is the correct way to divide by which ln?

The three basic rules are:

$\displaystyle log_a(b) + log_a(c) = log_a(b * c).$

$\displaystyle log_a(b) - log_a(c) = log_a\left(\dfrac{b}{c}\right).$

$\displaystyle b * log_a(c) = log_a\left(c^b\right).$