The question is to evaluate each logarithm:
My professor wants us to show work. Could you please tell me if I'm on the right track? Thank you!!!
1. log6^(1/36)= -2
2. ln(1/e^4)= 1/e^2
3. log7^1= 0
4. log4^64= ln64/ln4= 3
5. log10^8= ln8/ln10= .9031
6. log3^(radical 3)= I wasn't sure how to show the work on this one. I plugged it into my calculator and got .8263
#1
How did you get -2 ??
log 6^(1/36)
= (1/36) log 6 = appr .0216
#2
ln(1/e^4)
= ln1 - ln e^4
= 0 - 4lne
= 0 - 4(1) = -4
#3
log 7^1
= log 7 = appr .845
#4
log 4^64
= 64 log4 = appr 38.532
#5
log 10^8
= 8log10
= 8
#6 --- the only one you got right
Some of these blokes use
log6^(1/36) to mean log6(1/36), which is indeed -2
#2 ln(1/e^4) = ln(e^-4) = -4
#3 ok
#4 ok
#5 since the other logs specify a base, this must be just a common log, and log 10^8 = 8
#6 log3√3 = log33^(1/2) = 1/2
Hi thank you so much for helping me. Yeah, I am sorry. I do not know how to make the base numbers small like that. Thank you both again for the help!
Thanks Steve
Had an inkling that's what he was doing, but then it was contradicted in how he worked the last one.
btw, was it overkill in my solution to
http://www.jiskha.com/display.cgi?id=1461944692
Almost gave up on it, but found some clues on a webpage.
The subscripts are a trick of HTML. Most of the codes do not work on this site, but the sub and /sub tags work.
For superscripts, the sup and /sup tags work.
enclose the words in <> brackets to make them into HTML tags.
Sure! Let's go through each logarithm and check if your answers are correct.
1. log6^(1/36)
To evaluate this logarithm, you need to understand that the base of the logarithm is 6 and the exponent is 1/36. The answer should be the power to which the base must be raised to get the argument. In this case, it is asking what exponent we need to raise 6 to in order to get 1/36. Mathematically, it can be written as:
6^x = 1/36
To solve this equation, you need to find the exponent (x) that satisfies this equation. By taking the logarithm of both sides with base 6, we get:
log6 (6^x) = log6 (1/36)
x = log6 (1/36)
Now let's calculate it:
x = log6 (1/36)
x = log (1/36) / log (6)
Using a calculator, you can find:
x ≈ -2.
Therefore, your answer for 1. log6^(1/36) = -2 is correct.
2. ln(1/e^4)
In this logarithm, ln represents the natural logarithm with base e. The argument is 1/e^4. To solve this, you can simplify it as:
ln(1/e^4) = ln(1) - ln(e^4) = 0 - 4ln(e) = -4
Therefore, your answer for 2. ln(1/e^4) = 1/e^2 is incorrect. The correct answer should be -4.
3. log7^1
For this logarithm, the base is 7, and it is raised to the power of 1. The answer will be the power to which 7 must be raised to get 1. Since any number raised to the power of 0 is equal to 1, the logarithm of 1 to any base will always be 0.
Therefore, your answer for 3. log7^1 = 0 is correct.
4. log4^64
In this logarithm, the base is 4, and the exponent is 64. To evaluate it, you can use the change of base formula. Let's use the natural logarithm (ln) as the base:
log4^64 = ln(64) / ln(4)
Using a calculator, you can find:
log4^64 ≈ 3
Therefore, your answer for 4. log4^64 = ln64/ln4 = 3 is correct.
5. log10^8
Similar to the previous example, let's use the change of base formula with the natural logarithm as the base:
log10^8 = ln(8) / ln(10)
Using a calculator, you can find:
log10^8 ≈ 0.9031
Therefore, your answer for 5. log10^8 = ln8/ln10 = 0.9031 is correct.
6. log3^(sqrt(3))
In this case, the base is 3, and the exponent is the square root of 3. You mentioned that you plugged it into your calculator and got approximately 0.8263. If your calculator is set to compute logarithms with base 10, then it gives you the logarithm with base 10.
To get the logarithm with base 3, you can use the change of base formula:
log3^(sqrt(3)) = log(√3) / log(3)
Using a calculator, you can find:
log3^(sqrt(3)) ≈ 0.5493
Therefore, your answer for 6. log3^(sqrt(3)) = approximately 0.5493.
Overall, you did a good job on most of the logarithms. Just make sure to use the correct base and check the details in your calculations. Keep up the good work!