Help with logarithmic equations

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In summary, Homework Equations state that:logb(x)-logb(y) = logb(x/y)logb(x^n) = nlogb(x)andThe Attempt at a Solution states that:i am stuck on what to do because of the 4-2-2 portion. I've attempted it multiple times to get the solution (-4) but i just don't get how to do it because i keep getting the answer wrong.
  • #1
chirumu
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Homework Statement



[PLAIN]http://img225.imageshack.us/img225/6162/unledsxy.png

Homework Equations



logb(x)-logb(y) = logb(x/y)
logb(x^n) = nlogb(x)

The Attempt at a Solution



i am stuck on what to do because of the 4-2-2 portion. I've attempted it multiple times to get the solution (-4) but i just don't get how to do it because i keep getting the answer wrong.

I guess i don't understand how to combine it into one log this is my attempt nontheless:

log3(4-2)-2log3(6)
log3(2)-log3(12)
log3(2/12)
log3(x) = (2/12)

:/ I'm not sure what to do because i don't know how to treat the 4-2-2 :/ i know i can't combine the -2 where i want..

actually no wait, I've just had a mindwave. Would i convert 2 into a log as well.
I will try that now.
 
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  • #2
What's the question you're working on?

There is no question given in your post.
 
  • #3
chirumu said:

Homework Statement



[PLAIN]http://img225.imageshack.us/img225/6162/unledsxy.png[/QUOTE]

I am not certain what your question is, like Sammy, here. However, I tried to simplify the equation (quoted above) and it brought me to -4. I understand this is the answer you were looking for? If yes, read down:

Step 1: Figure out how to change "2" to a log of base 3.
Step 2: If you've managed to do step 1, then think of your laws of logarithm and then, indices.
 
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  • #4
Now that I've taken a closer look at your attempt to this problem, I see that you don't seem to know your laws of logarithm.

I'll try to do a breakdown of a few of the operations and hopefully, things will make more sense.

What you did was:

[itex]2\log_{3}6 = log_{3}(6\times2)[/itex]

Here's how it should have been manipulated:

[itex]2\log_{3}6 = log_{3}(6^2)= log_{3}36 [/itex]

Also, in this situation:

[itex]2 = log_{3}(3^2)[/itex]

I would urge you to review your laws of logarithm because it seems to me that this is not very clear to you. I think the website math2.org has some good explanations.
 
  • #5
thankyou for your help :)
i initially skipped this one and moved onto the others and after having done a few more its become more clear to me.

in the end:

log3(4)-2-2log3(6)

given a^c=b and loga(b)=c
to change 2 to base 3: log3(9) =2

so

log3(4)-log3(9)-2log3(6)

log3(4)-log3(9)-log3(6^2)

log3(4)-log3(9)-log3(36)


log3(4/9/36) = x

3(x) = (4/9/36)

x = -4

however i couldn't figure out -4 without a calculator :I I'm trying to stop using my calculator as much because its the devil. I've normally been able to solve the power, but in this instance the power is negative and the answer (4/9/36) is 0.01234... and i wasnt sure. Upon inspection given the fact that (4/9/36) IS 0.01234... i could kind of guess the power would be negative, but i would not of been able to figure it was -4. Could you offer any advice on this.
 
  • #6
Mépris said:
I am not certain what your question is, like Sammy, here. However, I tried to simplify the equation (quoted above) and it brought me to -4. I understand this is the answer you were looking for? If yes, read down:

The image in the OP is NOT an equation. An equation ALWAYS has "=" in it somewhere.

chirumu said:
thankyou for your help :)
i initially skipped this one and moved onto the others and after having done a few more its become more clear to me.

in the end:

log3(4)-2-2log3(6)

given a^c=b and loga(b)=c
to change 2 to base 3: log3(9) =2

so

log3(4)-log3(9)-2log3(6)

log3(4)-log3(9)-log3(6^2)

log3(4)-log3(9)-log3(36)


log3(4/9/36) = x

3(x) = (4/9/36)

x = -4
Your first post did not make this clear, but apparently you are supposed to simplify the given expression. When you rewrite an expression in a different form, use = to indicate that the different forms are equal.

You do not need to introduce a variable. Your work should look something like this:
log3(4) - 2 - 2log3(6)
= log3(4) - log3(9) - 2log3(6)
= log3(4/9) - 2log3(6)
= log3(4/9) - log3(36)
= log3(4/((9*36)

If you continue simplifying, you can get this down to a final result of -4.

chirumu said:
however i couldn't figure out -4 without a calculator :I I'm trying to stop using my calculator as much because its the devil. I've normally been able to solve the power, but in this instance the power is negative and the answer (4/9/36) is 0.01234... and i wasnt sure. Upon inspection given the fact that (4/9/36) IS 0.01234... i could kind of guess the power would be negative, but i would not of been able to figure it was -4. Could you offer any advice on this.
 

Related to Help with logarithmic equations

1. What is a logarithmic equation?

A logarithmic equation is an equation that contains a logarithm, which is a mathematical function that represents the inverse of an exponential function. It is used to solve for the power or exponent in an exponential equation.

2. How do I solve a logarithmic equation?

To solve a logarithmic equation, you first need to isolate the logarithm on one side of the equation. Then, you can use the properties of logarithms to simplify the equation and solve for the variable. Remember to always check your solution by plugging it back into the original equation.

3. What are the properties of logarithms?

The properties of logarithms are rules that can be used to simplify logarithmic equations. These include the product rule, quotient rule, power rule, and change of base rule.

4. How are logarithms used in real life?

Logarithms are used in a variety of fields, including science, engineering, finance, and computer science. They are used to solve exponential growth and decay problems, measure the intensity of earthquakes and sound, and calculate pH levels in chemistry.

5. What is the difference between natural logarithms and common logarithms?

Natural logarithms (ln) use the base e, which is an irrational number approximately equal to 2.718. Common logarithms (log) use the base 10. Both types of logarithms have their own set of rules and properties, but they can be converted to each other using the change of base rule.

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