Different ways of expressing logarithms

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In summary, to show that ^{2}log(e)=\frac{1}{ln2}, you can use the second relevant equation given in the homework, which states that ^{a}log(x) = ^{a}log(b)\cdot ^{b}log(x). By substituting log_b(x) = p and log_a(x) = q, you can prove the equation and show that ^{2}log(e)=\frac{1}{ln2}.
  • #1
Maxo
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Homework Statement


Show that [tex]^{2}log(e)=\frac{1}{ln2}[/tex]

Homework Equations


[tex]^{a}log(x) = ^{a}log(b)\cdot ^{b}log(x)[/tex][tex]^{a}log(x)=\frac{^{b}log(x)}{^{b}log(a)}[/tex]

The Attempt at a Solution


How can this be shown? I assume it can be done just using logarithm laws, but I don't see how. I tried manipulating around the factors, but I can't get to it. Can someone please show some way to show what the question asks.
 
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  • #2
I think you wrongly formatted [itex]log_a(x)[/itex] as [itex]a log(x)[/itex] instead.

You can use the second relevant equation you have noted above. What can you use as the value of 'b' in that equation to get something that could look like the RHS of your desired proof ? And what is [itex]log_k(k)[/itex] for any k in a valid domain?
 
  • #3
Infinitum said:
I think you wrongly formatted [itex]log_a(x)[/itex] as [itex]a log(x)[/itex] instead.
Where?

Infinitum said:
You can use the second relevant equation you have noted above. What can you use as the value of 'b' in that equation to get something that could look like the RHS of your desired proof ? And what is [itex]log_k(k)[/itex] for any k in a valid domain?
Actually I shouldn't have written that equation. I want to know how to do it, without using this equation. The point of the exercise is actually to make an example which leads to this equation. Is this possible?
 
  • #4
Maxo said:

Homework Statement


Show that [tex]^{2}log(e)=\frac{1}{ln2}[/tex]

Homework Equations


[tex]^{a}log(x) = ^{a}log(b)\cdot ^{b}log(x)[/tex][tex]^{a}log(x)=\frac{^{b}log(x)}{^{b}log(a)}[/tex]

The Attempt at a Solution


How can this be shown? I assume it can be done just using logarithm laws, but I don't see how. I tried manipulating around the factors, but I can't get to it. Can someone please show some way to show what the question asks.

The standard notation for the base-a logarithm of b is ##\log_a b## or ##\log_a(b)##, not ##{}^a \log b## or ##{}^a \log(b)##. Anyway, ##\log_a b = x## means that ##a^x = b##, so you should be able get everything you need from that.
 
  • #5
You could prove the relevant equation, or the question itself.

Let [itex]log_b(x) = p[/itex], and [itex]log_a(x) = q[/itex]
Substituting for x in the second equation, by using the definition of logarithm on the first, you should be able to prove the relevant equation.
 

Related to Different ways of expressing logarithms

1. What is the definition of a logarithm?

A logarithm is the inverse function of exponentiation, meaning it is the number that an exponent must be raised to in order to get a specific value. For example, the logarithm base 2 of 8 is 3, because 2 to the power of 3 is 8.

2. Why do we use logarithms?

Logarithms are useful for simplifying complex calculations involving large numbers or exponents. They also help to convert multiplication and division operations into addition and subtraction, making them easier to solve.

3. What are the different ways of expressing logarithms?

The most common ways of expressing logarithms are using the base 10 (common logarithm) or the base e (natural logarithm). Other ways include using a different base, such as base 2 or base 3, or using the notation "logb(x)" to represent "the logarithm base b of x".

4. How do we convert between different ways of expressing logarithms?

To convert from a logarithm with one base to another, we use the change of base formula: logb(x) = loga(x) / loga(b). For example, to convert from log base 2 to log base 10, we would divide the original log by log 2.

5. What are the properties of logarithms?

The main properties of logarithms include the product rule (log(ab) = log(a) + log(b)), the quotient rule (log(a/b) = log(a) - log(b)), and the power rule (log(ab) = b * log(a)). Another important property is the inverse relationship between logarithms and exponents, where logb(bx) = x and blogb(x) = x.

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