# Issue With Algebra of Logarithms

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• gibberingmouther
In summary, when working with logarithms, the order of operations matters. Exponentiation takes precedence over multiplication, and logarithms follow the same rules as exponents. In order to evaluate a logarithm with multiplication or division, the terms must be separated and evaluated separately before combining the results. This is a result of the PEMDAS (or BEDMAS) mnemonic, which emphasizes the order of operations in algebraic expressions. With practice, applying logarithm rules will become second nature.
gibberingmouther
I was just doing a homework problem that involved logarithms.

I noticed that order of operations matters when applying logarithm rules.

I'll use a different example from my homework problem to illustrate what I'm talking about.

ln(5*2^3) does not equal 3*(ln(5*2))

Apparently you have to do ln5+3*ln2 - you have to separate out the products before multiplying the second term by the exponent! It kind of makes sense but I can't say exactly why this is how it has to be.

Can anyone confirm that what I have said is true and explain why?

Logarithms as numbers, are exponents, and they follow the rules of exponents.

Not know which base you want to work with, but log(5*2^3)
log(5)+log(2^3)
log(5)+3*log(2)
and your book would explain how and why better than I can right now.

jedishrfu
Well consider the simpler example of ##5*2^3## vs ##(5*2)^3##

mathwonk and gibberingmouther
jedishrfu said:
Well consider the simpler example of ##5*2^3## vs ##(5*2)^3##
That is 40.
If want log(40), this is if base ten,
log(4*10)
log(4)+log(10)
log(4)+1
0.60206+1
1.60206, approximately

This also means that 10^1.60206=40, approximately
101.60206=40

gibberingmouther and jedishrfu
jedishrfu said:
Well consider the simpler example of ##5*2^3## vs ##(5*2)^3##
Yes, that does put things in perspective a bit!

This can also help reenforce things a little more.

23*25=2*2*2*2*2*2*2*2
2(3+5)
28The EXPONENTS are LOGARITHMS. Here, the base was 2.

jedishrfu
gibberingmouther said:
ln(5*2^3) does not equal 3*(ln(5*2))
There's an order of operations at play here that has nothing to do with logarithms.

In the expression 5 * 2^3, the exponent operator takes precedence over the multiplication, making this expression equal to 5 * 8. The mnemonic is PEMDAS (or BEDMAS), with each emphasizing that exponentiation takes precedence over multiplication or division. (he latter two operations have the same precedence.)

Going back to the log problem, this means that
##\log(5 \cdot 2^3) = \log(5 \cdot 8) = \log(5) + \log(8)##

If the log expression had been ##\log((5\cdot 2)^3)##, then you could have used the log property you're using to get ##3\log(5 \cdot 2) = 3\log(10)##

gibberingmouther
Mark44 said:
There's an order of operations at play here that has nothing to do with logarithms.

In the expression 5 * 2^3, the exponent operator takes precedence over the multiplication, making this expression equal to 5 * 8. The mnemonic is PEMDAS (or BEDMAS), with each emphasizing that exponentiation takes precedence over multiplication or division. (he latter two operations have the same precedence.)

Going back to the log problem, this means that
##\log(5 \cdot 2^3) = \log(5 \cdot 8) = \log(5) + \log(8)##

If the log expression had been ##\log((5\cdot 2)^3)##, then you could have used the log property you're using to get ##3\log(5 \cdot 2) = 3\log(10)##

Okay, that helps a lot. I like to be able to do the algebra following the rules for it without having to memorize special exceptions. Here, this is the way to look at it from the perspective of PEMDAS. Thanks!

Learning the rules for logarithms should not be very long in time. Really only a few of them are there. Applying them in steps will become part of your nature.

## 1. What is algebra of logarithms?

The algebra of logarithms is a set of rules and properties used to manipulate logarithmic expressions in algebraic equations. It is based on the relationship between logarithms and exponents, and allows for simplification and solving of equations involving logarithms.

## 2. What are the common issues with algebra of logarithms?

One common issue is when the base of the logarithm is not specified, as different bases result in different values. Another issue is when the logarithmic expression contains a variable, making it difficult to solve algebraically. Additionally, using improper logarithmic properties can lead to errors in the solution.

## 3. How do I solve equations involving logarithms?

The first step is to simplify the logarithmic expression using the properties of logarithms. Then, isolate the logarithmic term on one side of the equation and exponentiate both sides using the inverse of logarithms. Finally, solve for the variable using algebraic techniques.

## 4. Can logarithmic equations have more than one solution?

Yes, logarithmic equations can have multiple solutions. This can happen when the logarithmic expression has a base greater than 1, resulting in multiple values that satisfy the equation. It is important to check for extraneous solutions, which are values that do not actually satisfy the original equation.

## 5. How is algebra of logarithms used in real life?

Algebra of logarithms is used in fields such as science, finance, and engineering to model and solve various problems. In science, it is used to analyze exponential growth and decay, while in finance it is used to calculate compound interest. In engineering, it is used to model and solve various exponential and logarithmic relationships in systems and processes.

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