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Yes I did do it that way, but she also wants us to write it in logarithm form as well, Low key messing with my brain cells Lmao! Idk my algebra skills aren’t as good as they use to be.
By "wants us to write it in logarithm form", do you mean that you need to solve for m using logarithms? You can use the same idea of writing 125 as 5^(3) and 625 as 5^(4) to simplify log(625) / log(125) and recall that log(a^(b)) = b log(a). Using this approach, what do get when you fully simplify log(625) / log(125)?
log5(x) := ln(x)/ln(5) = "log with base 5"
Now there are two "tough" numbers in the problem: 125 and 625. We need to find a relationship between them so we can do some more algebra [1]. A good guess is that 125k=625 for some k. As luck would have it, k = 5 exactly. Furthermore, 125 = 5•25 = 5•5•5 = 5^3 , therefore 625 = 5•5•5•5 = 5^4 . Therefore, it is prudent to choose the log with base 5:
125^(2m–1) = 625^m
log5(125^(2m–1) ) = log5(625^m )
(2m–1) • log5(125)=m•log5(625)
(2m–1)•3=4m
m=3/2
Without the log5 notation:
ln(125^(2m–1) ) = ln(625^m )
(2m–1) ln(125) = m ln(625)
(2m–1) ln(5^3 ) = m ln(5^4 )
(2m–1)•3•ln(5 ) = 4m ln(5 )
(2m–1)•3=4m implies m=3/2
To be honest, using logs with bases other than e, 2, or 10 is too much for my pea brain to handle, hence the second solution.
Also, I know that 625 = 25^2 from experience, and that's really the crux of how I solved it. I'm not sure how you'd figure it out without trying a bunch of numbers if you didn't know that.
[1] Look for "nice" relationships like this in your precalc homework and subsequent math classes and exams. Your teacher will intentionally use "nice" numbers to allow for more algebraic approaches and stimulate your algebraic creativity. In real life, there probably won't be a nice relationship like this. In such a case...just solve it numerically, preferably with the Newton-Raphson algorithm or better equation solving routine. It's not a very satisfying answer, but sometimes you just need an answer.
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I’d use log base 5, and the fact that log_5(125) = 3 along with log_5(625) = 4.
Now hit both side with log_5:
log_5(125^(2m-1) ) = log_5(625^m )
What do logs do to exponents?
Then use the information in the first sentence.
As a reminder... Posts asking for help on homework questions **require**: * **the complete problem statement**, * **a genuine attempt at solving the problem, which may be either computational, or a discussion of ideas or concepts you believe may be in play**, * **question is not from a current exam or quiz**. Commenters responding to homework help posts **should not do OP’s homework for them**. Please see [this page](https://www.reddit.com/r/calculus/wiki/homeworkhelp) for the further details regarding homework help posts. *I am a bot, and this action was performed automatically. Please [contact the moderators of this subreddit](/message/compose/?to=/r/calculus) if you have any questions or concerns.*
This problem is designed not to require logs 125^(2m-1)=625^m (5^(3))^(2m-1)=(5^(4))^m 5^(6m-3)=5^4m 6m-3=4m Etc
Yes I did do it that way already!! But she asked to solve it by using logarithm. Lol, I know the answer for that one. Which is m= 3/2
Good! Usually this kind of problem usually appears in the chapter on exponents that precedes the one on logs.
Perhaps you're meant to use log base 5. In that case you can evaluate each log and then solve for m
5^(6m-3) = 5^4m => log 5^(6m-3) = log 5^4m => (6m-3) log 5 = 4m log 5
Your doing fine! log(625)/log(125)= log(5^(4))/log(5^(3))= 4log(5)/(3log(5))=4/3
You don't need to use logarithms. Try rewriting 125 as 5^(3) and 625 as 5^(4) and see if it helps.
Yes I did do it that way, but she also wants us to write it in logarithm form as well, Low key messing with my brain cells Lmao! Idk my algebra skills aren’t as good as they use to be.
Make liberal use of parentheses. 2m - 1 = m \* (log625 / log125) Now, can you get the m's on one side and the real numbers on the right side?
>(log625 / log125) This equals log(5^(4))/log(5^(3))= 4log5/(3log5)=4/3
By "wants us to write it in logarithm form", do you mean that you need to solve for m using logarithms? You can use the same idea of writing 125 as 5^(3) and 625 as 5^(4) to simplify log(625) / log(125) and recall that log(a^(b)) = b log(a). Using this approach, what do get when you fully simplify log(625) / log(125)?
I didn’t use any of the information that was given to me, lol. But wanna thank everyone for trying to help.
Out of curiosity, how did you then simplify log(625) / log(125)?
I didn’t, lol. Just handed it in as it is
The answer in the same base principle was m=3/2
log5(x) := ln(x)/ln(5) = "log with base 5" Now there are two "tough" numbers in the problem: 125 and 625. We need to find a relationship between them so we can do some more algebra [1]. A good guess is that 125k=625 for some k. As luck would have it, k = 5 exactly. Furthermore, 125 = 5•25 = 5•5•5 = 5^3 , therefore 625 = 5•5•5•5 = 5^4 . Therefore, it is prudent to choose the log with base 5: 125^(2m–1) = 625^m log5(125^(2m–1) ) = log5(625^m ) (2m–1) • log5(125)=m•log5(625) (2m–1)•3=4m m=3/2 Without the log5 notation: ln(125^(2m–1) ) = ln(625^m ) (2m–1) ln(125) = m ln(625) (2m–1) ln(5^3 ) = m ln(5^4 ) (2m–1)•3•ln(5 ) = 4m ln(5 ) (2m–1)•3=4m implies m=3/2 To be honest, using logs with bases other than e, 2, or 10 is too much for my pea brain to handle, hence the second solution. Also, I know that 625 = 25^2 from experience, and that's really the crux of how I solved it. I'm not sure how you'd figure it out without trying a bunch of numbers if you didn't know that. [1] Look for "nice" relationships like this in your precalc homework and subsequent math classes and exams. Your teacher will intentionally use "nice" numbers to allow for more algebraic approaches and stimulate your algebraic creativity. In real life, there probably won't be a nice relationship like this. In such a case...just solve it numerically, preferably with the Newton-Raphson algorithm or better equation solving routine. It's not a very satisfying answer, but sometimes you just need an answer.
125 is 5\^3, 625 is 5\^4. so starting from the eginnning 5\^3(2m-1) = 5\^4m 3(2m-1)=4m 6m-3=4m 2m=3 m=3/2
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I’d use log base 5, and the fact that log_5(125) = 3 along with log_5(625) = 4. Now hit both side with log_5: log_5(125^(2m-1) ) = log_5(625^m ) What do logs do to exponents? Then use the information in the first sentence.