Solving Logarithm Equations ( and Changing Forms)

[tristatefabricatorsinc]

I am having trouble with a few problems, maybe someone can help me...

Find the following value, round to the nearest hundredths place...

log sub2 6

(Sorry, I am not sure how to say log sub a number)

Is there a way to plug this into the calculator? Or do I have to break it down some how?

Second question is...

Solve for x: x = log sub3 1 / 81

I am unsure where to begin here...?

 

[Unco]

Let \(\displaystyle \mbox{ y = \log_2{(6)}\)

Take the exponential base 2 of both sides:

\(\displaystyle \mbox{ 2^y = 6}\)

Take logs base 10 of both sides and solve for y, that is \(\displaystyle \mbox{\log_2{(6)}}\), again:

\(\displaystyle \mbox{ y = \frac{\log_{10}{(6)}}{\log_{10}{(2)}}}\)

You can evaluate this with your calculator. (Your calculator probably has log base e too.)

The second one is not dissimilar.

 

[Mrspi]

I am having trouble with a few problems, maybe someone can help me...

Second question is...

Solve for x: x = log sub3 1 / 81

I am unsure where to begin here...?

Start by changing to exponential form. If log<SUB>b</SUB> a = m, then
b<SUP>m</SUP> = a

If log<SUB>3</SUB> (1/81) = x, then
3<SUP>x</SUP> = 1/81

Now, let's see if we can do something with 1/81. I note that 81 is a power of 3:
1/81 = 1/3<SUP>4</SUP>, or 3<SUP>-4</SUP>, so

3<SUP>x</SUP> = 3<SUP>-4</SUP>

The bases are the same, and the expressions are equal. What can you conclude about the exponents?

I hope this helps you.

 

[tristatefabricatorsinc]

Hello,

I understand how Mrspi got to the end of the first question. The only thing I do not understand is what to do with the exponents. Am I supposed to use one of the rules of exponents to finish it?

For the second part of my question, I am unsure how you are able to just take the 2 from the right side and move it to the left, and then I am really confused how a fraction came out of this. Is there another way of explaining it in a bit more detail?

All the help is greatly appreciated![/tex][/url][/list][/code][/quote]

 

[Mrspi]

Hello,

I understand how Mrspi got to the end of the first question. The only thing I do not understand is what to do with the exponents. Am I supposed to use one of the rules of exponents to finish it?

All the help is greatly appreciated![/tex][/url][/list][/code]

[/quote]

Maybe you are making this harder than it needs to be.

If b<SUP>m</SUP> = b<SUP>n</SUP>, then it must be true that m = n.

If 3<SUP>x</SUP> = 3<SUP>-4</SUP>, then ????????

 

[tristatefabricatorsinc]

The exponents are the same?

 

[Mrspi]

Hello,


For the second part of my question, I am unsure how you are able to just take the 2 from the right side and move it to the left, and then I am really confused how a fraction came out of this. Is there another way of explaining it in a bit more detail?

All the help is greatly appreciated![/tex][/url][/list][/code]

[/quote]

Unco showed you this:
let y = log<SUB>2</SUB> 6

Change to exponential form:
2<SUP>y</SUP> = 6

Take the log (and I'm using base 10 logs here) of both sides:
log 2<SUP>y</SUP> = log 6

Use the rule of logs which says that log b<SUP>a</SUP> = a log b

y log 2 = log 6
You're trying to solve for y.....wouldn't your next step be to divide both sides by log 2? Do you see where the fraction comes from now?

 

[tristatefabricatorsinc]

3^x = (1/81)

Well, 3^4 is 81, so the answer must be -4

3^-4 = (1/81)

Is this correct?

 

[tkhunny]

That's nice, but what are you going to do when it gets harder and you don't see it.

3^x = 1/2187

You need to learn to manipulate the exponential/logarithmic expressions and equations.