Solving log b (1/4) = -1/2; evaluating log 5 cube root of 25
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There are usually a couple of ways to look at these problems. For example, for the first problem, you can start by raising both sides to the power \(\displaystyle b\). For the second problem, remember that a cube root is the same as the \(\displaystyle 1/3\) power. Also, review the laws of logarithms for both problems.
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Remember the relationship between logs and exponents ...
Hi Alice:
Your statement above leads me to suspect that you don't really understand the relationship between logarithms and exponents. You should follow Roy's advice to REVIEW the definitions and properties. (You should also have a good working knowledge of the properties of exponents, too.)
log[sub:2isyujxt]b[/sub:2isyujxt](expression) = x means that b[sup:2isyujxt]x[/sup:2isyujxt] = expression (and vice versa!)
When we see log[sub:2isyujxt]b[/sub:2isyujxt](expression) = 5
we realize this log statement is telling us that SOME EXPONENT IS 5.
Specifically, 5 is the power to which b must be raised in order for this power to equal the "expression".
b[sup:2isyujxt]5[/sup:2isyujxt] = expression
Once you remember this, then you realize that it's possible to rewrite the statement as a power instead of a log.
(When working with logarithms, it's often necessary to switch back and forth between the logarithmic statement and the corresponding exponential statement.)
With the two exercises you posted, if you rewrite the log statements as above, then you should be able to solve both b and x by inspection.
(1) log[sub:2isyujxt]b[/sub:2isyujxt](1/4) = -1/2
(2) log[sub:2isyujxt]5[/sub:2isyujxt](25[sup:2isyujxt]1/3[/sup:2isyujxt]) = x
Here are the examples you asked for.
For the first exercise ...
log[sub:2isyujxt]k[/sub:2isyujxt](1/81) = -4
This log statement defines that k must be raised to the power of -4 in order to equal 1/81. Rewrite as an exponential statement using the relationship as defined above:
k[sup:2isyujxt]-4[/sup:2isyujxt] = 1/81
Since k[sup:2isyujxt]-4[/sup:2isyujxt] is the same as 1/k[sup:2isyujxt]4[/sup:2isyujxt], we have:
1/k[sup:2isyujxt]4[/sup:2isyujxt] = 1/81
k[sup:2isyujxt]4[/sup:2isyujxt] = 81
k = 3
For the second exercise ...
log[sub:2isyujxt]2[/sub:2isyujxt](16[sup:2isyujxt]1/5[/sup:2isyujxt]) = x
Again, the definition of logs tells us that 2 must be raised to the power of x in order to equal the fifth root of 16. Rewrite as an exponential statement using the definition of log[sub:2isyujxt]2[/sub:2isyujxt]:
2[sup:2isyujxt]x[/sup:2isyujxt] = 16[sup:2isyujxt]1/5[/sup:2isyujxt]
Since 16 is the same as 2[sup:2isyujxt]4[/sup:2isyujxt], we have:
2[sup:2isyujxt]x[/sup:2isyujxt] = (2[sup:2isyujxt]4[/sup:2isyujxt])[sup:2isyujxt]1/5[/sup:2isyujxt]
2[sup:2isyujxt]x[/sup:2isyujxt] = 2[sup:2isyujxt]4/5[/sup:2isyujxt]
x = 4/5
Try to understand the relationship; if you simply look for patterns between examples and exercises, followed by blindly matching numbers with "templates", then you're likely to continue to hate logs because you won't know what to do when a template example is not available.
Let us know if you still need help. Please show your work, too.
Cheers,
~ Mark
alice said:
Hi Alice:
Your statement above leads me to suspect that you don't really understand the relationship between logarithms and exponents. You should follow Roy's advice to REVIEW the definitions and properties. (You should also have a good working knowledge of the properties of exponents, too.)
log[sub:2isyujxt]b[/sub:2isyujxt](expression) = x means that b[sup:2isyujxt]x[/sup:2isyujxt] = expression (and vice versa!)
When we see log[sub:2isyujxt]b[/sub:2isyujxt](expression) = 5
we realize this log statement is telling us that SOME EXPONENT IS 5.
Specifically, 5 is the power to which b must be raised in order for this power to equal the "expression".
b[sup:2isyujxt]5[/sup:2isyujxt] = expression
Once you remember this, then you realize that it's possible to rewrite the statement as a power instead of a log.
(When working with logarithms, it's often necessary to switch back and forth between the logarithmic statement and the corresponding exponential statement.)
With the two exercises you posted, if you rewrite the log statements as above, then you should be able to solve both b and x by inspection.
(1) log[sub:2isyujxt]b[/sub:2isyujxt](1/4) = -1/2
(2) log[sub:2isyujxt]5[/sub:2isyujxt](25[sup:2isyujxt]1/3[/sup:2isyujxt]) = x
Here are the examples you asked for.
For the first exercise ...
log[sub:2isyujxt]k[/sub:2isyujxt](1/81) = -4
This log statement defines that k must be raised to the power of -4 in order to equal 1/81. Rewrite as an exponential statement using the relationship as defined above:
k[sup:2isyujxt]-4[/sup:2isyujxt] = 1/81
Since k[sup:2isyujxt]-4[/sup:2isyujxt] is the same as 1/k[sup:2isyujxt]4[/sup:2isyujxt], we have:
1/k[sup:2isyujxt]4[/sup:2isyujxt] = 1/81
k[sup:2isyujxt]4[/sup:2isyujxt] = 81
k = 3
For the second exercise ...
log[sub:2isyujxt]2[/sub:2isyujxt](16[sup:2isyujxt]1/5[/sup:2isyujxt]) = x
Again, the definition of logs tells us that 2 must be raised to the power of x in order to equal the fifth root of 16. Rewrite as an exponential statement using the definition of log[sub:2isyujxt]2[/sub:2isyujxt]:
2[sup:2isyujxt]x[/sup:2isyujxt] = 16[sup:2isyujxt]1/5[/sup:2isyujxt]
Since 16 is the same as 2[sup:2isyujxt]4[/sup:2isyujxt], we have:
2[sup:2isyujxt]x[/sup:2isyujxt] = (2[sup:2isyujxt]4[/sup:2isyujxt])[sup:2isyujxt]1/5[/sup:2isyujxt]
2[sup:2isyujxt]x[/sup:2isyujxt] = 2[sup:2isyujxt]4/5[/sup:2isyujxt]
x = 4/5
Try to understand the relationship; if you simply look for patterns between examples and exercises, followed by blindly matching numbers with "templates", then you're likely to continue to hate logs because you won't know what to do when a template example is not available.
Let us know if you still need help. Please show your work, too.
Cheers,
~ Mark
mmm4444bot
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alice said:
Hello Alice:
If you are unsure of your result, then why not double-check it? In fact, it's a good idea to double-check final results everytime you can make time.
Your answer says that 8 is the base that needs to be raised to the power of -1/2 in order to equal 1/4.
Is this true?
8[sup:9t6otitj]-1/2[/sup:9t6otitj] = 1/4
If this is true, then your answer is correct. If this is not true, then your answer is not correct.
~ Mark
mmm4444bot
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Review the Properties.
Hello:
Roy and I both suggested that you review the properties of logarithms. Have you done that, yet?
On this last problem, use the properties of logs to rewrite the LHS (left-hand side) of the equation as a single log.
Then you can equate the arguments appearing in the log function on each side. For example, once you get something that is log=log, you can do this:
log(4x^2 + 2) = log(10x)
4x^2 + 2 = 10x
Viola! No more logs.
If you need help (after reveiwing the properties of logs) consolidating the LHS into a single logarithm, then please post your work.
Cheers,
~ Mark
PS: Please post new exerices as a new discussion; otherwise, we may not notice them at the end of prior discussions. BTW, did you check your previous answer for b?
Hello:
Roy and I both suggested that you review the properties of logarithms. Have you done that, yet?
On this last problem, use the properties of logs to rewrite the LHS (left-hand side) of the equation as a single log.
Then you can equate the arguments appearing in the log function on each side. For example, once you get something that is log=log, you can do this:
log(4x^2 + 2) = log(10x)
4x^2 + 2 = 10x
Viola! No more logs.
If you need help (after reveiwing the properties of logs) consolidating the LHS into a single logarithm, then please post your work.
Cheers,
~ Mark
PS: Please post new exerices as a new discussion; otherwise, we may not notice them at the end of prior discussions. BTW, did you check your previous answer for b?
i used
log[subza1d1bi]b[/subza1d1bi]x - log[subza1d1bi]b[/subza1d1bi] y = log [subza1d1bi]b[/subza1d1bi] x/y
2 log[subza1d1bi]2[/subza1d1bi] (3/x) = log[subza1d1bi]2[/subza1d1bi] 45
2* 3/x = 45
6 = 45x
6/45 = x
I really don't understand unfortunately. I have a lot of trouble with these. I have read and reread the properties and can not grasp it. Is there an example that can help?? The examples before were very helpful thank you!!
Alice
log[sub
za1d1bi]b[/subza1d1bi]x - log[subza1d1bi]b[/subza1d1bi] y = log [subza1d1bi]b[/subza1d1bi] x/y2 log[sub
za1d1bi]2[/subza1d1bi] (3/x) = log[subza1d1bi]2[/subza1d1bi] 452* 3/x = 45
6 = 45x
6/45 = x
I really don't understand unfortunately. I have a lot of trouble with these. I have read and reread the properties and can not grasp it. Is there an example that can help?? The examples before were very helpful thank you!!
Alice
mmm4444bot
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alice said:
Whoops, you missed one of the Laws of Logarithms:
a * log(x) = log(x^a)
In other words, when you have a log multiplied by a coefficient, then you may move the coefficient to the exponent position as shown above.
If you can remember that logs are exponents, then this makes sense.
Consider the following.
log(x) = b
This statement tells us that b is an exponent.
Naturally, if we see 2 * log(x), then that tells us that the exponent is being multiplied by 2.
So, instead of 10^b = x, we have 10^(2b) = x.
There are only three Laws of Logarithms. Memorize them.
There is only one reason why you don't yet grasp how to work with these types of equations. You simply have not done enough exercises. So, hang in there! After you've been exposed to enough exercises, you will start to recognize the patterns and relationships. If you still struggle after completing your homework, then you need to do extra homework.
It's easy for students to put away the books when the assignments are finished. It's not so easy for students to find self-discipline AND resolve to do as many extra problems as is necessary to reach a level of self-confidence.
When I was in college, the professor may have assigned about 20 exercises daily. I usually ended up having to do 50 exercises a day before I felt that I was skilled enough to take an exam. It felt good when I finally saw the "big" picture.
You are the only person who knows when you're skillful enough to pass an exam. Try to work as many exercises as possible; the "big" picture will eventually develop in your mind.
Cheers,
~ Mark
mmm4444bot
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It looks like you don't want to think very much ...
I'm repeating my earlier post regarding how to check one's work!
I'm repeating my earlier post regarding how to check one's work!
mmm4444bot said:alice said:
Hello Alice:
If you are unsure of your result, then why not double-check it? In fact, it's a good idea to double-check final results everytime you can make time.
Your answer says that 8 is the base that needs to be raised to the power of -1/2 in order to equal 1/4.
Is this true?
8[sup:2jmf2s6i]-1/2[/sup:2jmf2s6i] = 1/4
If this is true, then your answer is correct. If this is not true, then your answer is not correct.
~ Mark
mmm4444bot
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Hi Alice:
Whenever you check a final result, you need to use the original.
The original equation is
2 * log[sub:2dsqkmqy]2[/sub:2dsqkmqy](3) - log[sub:2dsqkmqy]2[/sub:2dsqkmqy](x) = log[sub:2dsqkmqy]2[/sub:2dsqkmqy](45)
If you substitute 5 for x, then are the two sides of the equation equal?
~ Mark
My edit is adding the following note: I keep forgetting that most people don't have a log base 2 button on their calculator! Sorry about that. I just posted more information, but you need to move to page 2 of this discussion to see it, Alice...
Whenever you check a final result, you need to use the original.
The original equation is
2 * log[sub:2dsqkmqy]2[/sub:2dsqkmqy](3) - log[sub:2dsqkmqy]2[/sub:2dsqkmqy](x) = log[sub:2dsqkmqy]2[/sub:2dsqkmqy](45)
If you substitute 5 for x, then are the two sides of the equation equal?
~ Mark
My edit is adding the following note: I keep forgetting that most people don't have a log base 2 button on their calculator! Sorry about that. I just posted more information, but you need to move to page 2 of this discussion to see it, Alice...
mmm4444bot
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Oops!
Oh, I suppose it would help if I explained how to use a calculator to find log base 2.
The LOG button on your calculator is base 10 only.
Use the following formula to determine logs for bases other than 10.
log[sub:e8gng4yv]b[/sub:e8gng4yv](x) = log[sub:e8gng4yv]10[/sub:e8gng4yv](x) / log[sub:e8gng4yv]10[/sub:e8gng4yv](b)
If you want the calculator to give you the value of log[sub:e8gng4yv]2[/sub:e8gng4yv](45), then the formula goes like this:
log[sub:e8gng4yv]2[/sub:e8gng4yv](45) = log[sub:e8gng4yv]10[/sub:e8gng4yv](45) / log[sub:e8gng4yv]10[/sub:e8gng4yv](2)
so you would enter the log base 2 of 45 into the calculator as LOG 45 / LOG 2.
The calculator should display 1.5849...
Now, you should be able to use a calculator to check your previous answer of x = 5.
Oh, I suppose it would help if I explained how to use a calculator to find log base 2.
The LOG button on your calculator is base 10 only.
Use the following formula to determine logs for bases other than 10.
log[sub:e8gng4yv]b[/sub:e8gng4yv](x) = log[sub:e8gng4yv]10[/sub:e8gng4yv](x) / log[sub:e8gng4yv]10[/sub:e8gng4yv](b)
If you want the calculator to give you the value of log[sub:e8gng4yv]2[/sub:e8gng4yv](45), then the formula goes like this:
log[sub:e8gng4yv]2[/sub:e8gng4yv](45) = log[sub:e8gng4yv]10[/sub:e8gng4yv](45) / log[sub:e8gng4yv]10[/sub:e8gng4yv](2)
so you would enter the log base 2 of 45 into the calculator as LOG 45 / LOG 2.
The calculator should display 1.5849...
Now, you should be able to use a calculator to check your previous answer of x = 5.