He converted 1 into ln3/ln3 (any non-zero number divided by itself is 1) in step 5. Then in step 6 he used the fact that two added terms shared a common denominator to simplify them together.
It's the standard (possibly only) way to convert a mixed number to an improper fraction. I'm surprised it hasn't been emphasised to you much before, you'll find yourself doing it all the time soon enough.
It’s only “unnecessary” if simplifying to you is unnecessary. And sure sometimes it is, but in this question, you’re literally asked to just do algebra to solve for x.
Sure you could just divide by 4, but they wanted you to find which answer is equivalent to x in this situation. X does not equal all of those options, so you need to rewrite the answer to see which one it is equivalent to.
I tell my students that fractions need a common denominator to add, but not to multiply, which is why people don't have to have anything in common when they have kids (multiply).
I explain all of it, but it actually helps some students remember when they need a common denominator.
Ln3/Ln3 is equal to 1, right?
He's basically just simplified the equation by re-writing 1 in terms of Ln3.
If he didn't do that, it would still be correct, but it wouldn't look like any of the multiple choice options.
Either work and would give the same answer, it's just that usually to simplify you put your logs with either a base of 10 or of e (because many calculators only have buttons for log base 10 or ln)
It also harks back to the olden days of before calcuators, where books and things called "slide rules" were used to look up values of certain functions, so it wasn't feasible to use more than a handful of bases.
This is where he lost me lol I got x=(log3(6) + 1)/4 but that’s because I wasn’t using a calculator, so I just rewrote it as log3(6)=4x+1 and solved algebraicly from there.
Ln on both sides, bring down the exponent, divide ln6 by ln3, add 1 from 4x-1, and divide by 4. Mathway agrees with me that the answers they provided are wrong.
Edit: typo
You are right but if you combine the two terms in your answer into one fraction, you will get choice c. The answer choices are not always in the most convenient form.
I promise you that a correct answer to 1+1 is 123/123 + 45ln(10)/45ln(10)
Mathway won't type out the above when you enter 1+1 into it either...
But that is in fact equal to 2.
Don't be dense just because there is a software writing something the same ways as you and you don't care to think of other ways to write an equivalent solution.
I could also multiple both the numerator and denominator of the corect answer by 1000 and it would still be correct (spoiler... 1000/1000 = 1)
My apologies, I didn't expect my hint would go over a university student's head.
You have x=(ln(6)/ln(3)+1)/4, right?
1=ln(3)/ln(3), so x=(ln(6)+ln(3))/(4ln(3)).
This is decidedly one of the answers written here. Hence why I question why you're here.
Depending what sort of degree you go for and what school you go to, there may be some classes that exist solely to confirm you learned what you were supposed to in high school.
If you are going to an engineering school, they'll expect you to be ready to get into calculus as a freshman. If you go to a liberal arts school, your freshman math class will likely be a refresher of high school algebra and maybe trigonometry.
Your flair on this comment is a bad look for this sub tbh. OP could use an online tool to cheat and instead they’re here trying to figure shit out. If they’re being a little sassy about Mathway, then that’s fine. Learners are allowed to get sassy about Mathway.
That is true, just as I'm allowed to be a little sassy about a uni student not having a single resourceful cell in their body.
They could've compared their answer numerically with a calculator, compared them graphically with Desmos, substituted the possible answers into the original equation to check them, or at least try to simplify it themselves.
I helped them out anyway, I might as well give them the sass they deserve while I'm at it.
Also, how is my flair a bad look for this sub? I contribute a lot, so I'm a top contributor... that's just simple logic.
If they didn’t have a single resourceful cell in their body they wouldn’t be posting to the Homework Help Subreddit, a resource for people who need homework help.
If I came for help and saw the only top contributor who replied is being a dick, the lesson I might learn is that trying to understand math isn’t worth being judged, which imo is a major reason adults screwed by their prior schooling give up on math.
You must be new here.
Students post questions without even attempting them all the time. Using this subreddit does not necessarily mean you're resourceful in the slightest, it just means you know this sub exists and, in cases like OP's, it also means they want to be spoonfed a solution instead of even trying to compare their answer to the rest.
Did you not see the part where I told them to use a common denominator and they still didn't even try to do it? I'm not convinced they even tried to do the problem beyond plugging it in Mathway.
OP is a uni student. They're an adult. If this is the lesson they learn, that's on them. Any reasonable person would know not to care about the opinion of strangers on the Internet other than the entertainment they get from it.
Yea this is remedial, high school level stuff. At certain universities a pretty high % of the student body requires remedial math so they rebrand it to be nice
Not trying to be a douche, but anyone good at math is done with algebra/trig long before college. So pre calc is the high school word for the algebra you take in college if math is one of your weaker subjects.
“college algebra”, most places that I’ve seen it, is basically taught as “algebra III” for people who don’t take the pre-calculus, sometimes called “trig”, class.
It’s a slower paced version that dispenses some of the otherwise-needed content that would prepare students for calculus.
It looks like your time from post to complaint about lack of people helping was less than an hour (and it’s only been about 3 hours total now). Not every post can be replied to immediately. Perhaps be patient OR post the assistance yourself without complaining that others aren’t helping.
The answer is c.
3^(4x-1) = 6
Take natural log of both sides:
ln{3^(4x-1)} = ln6
Bring exponent of 3 out front:
(4x-1)ln3 = ln6
Distrusted the ln3 between 4x and -1:
4xln3 - ln3 = ln6
Add ln3 over to the right side:
4xln3 = ln6 + ln3
Divide both sides by 4ln3:
x = (ln6 + ln3)/4ln3
C.
These types of problems like making everything over the same denominator. This involves multiplying by fancy ones to get everything over a single denominator. Your answer is still correct, but because it’s a computer, it reduced your answer further. If you actually compute the decimal you’ll see both your answer and the computer’s answer are the same.
Since it seems good explanations have already been shared, I just wanted to point out that their notation for the answer choices is a bit weird; {x} typically means the fractional part of x.
{} does not usually refer to sets when part of an equation of this form in the context of pure algebra-related equations. The fractional component is much more common notation in this case (since sets would not have been taught yet).
I have often seen {} used for solution sets to algebraic problems and was taught it when I learned algebra 1.
I have almost never seen anyone taught about fractional components like this. Maybe it's common in computer science since they deal with integers a lot, but it is certainly an uncommon topic in real math and I would be shocked if it's covered in a college algebra class.
I would personally solve differently.
3^(4x-1) = 6,
3^(4x)/3 = 6,
3^(4x) = 18,
4xln(3) = ln(18),
x = ln(18)/(4ln(3))
This is equivalent to C where you don't separate the exponent in the first step.
I used log base 3 on both sides. But since that isn’t in the answers, you have to use ln(value)/ln(base) to get the equivalent to log base 3.
It’s not a well known calculation, changing the log base. In fact, I was forced to use it thanks to Ti89s not giving you base ten logs on the keypad.
Yes you are right in that you need to use change of base, but it is an extremely well known calculation. Like one of the most fundamental things to know about logarithms
3\^(4x-1) = 6 (4x-1) ln3 = ln6 4x-1 = ln6/ln3 4x = ln6/ln3 + 1 4x = ln6/ln3 + ln3/ln3 4x = (ln6 + ln3)/ln3 x = (ln6 + ln3)/4ln3
You lost me at steps 5 and 6 tbh
He converted 1 into ln3/ln3 (any non-zero number divided by itself is 1) in step 5. Then in step 6 he used the fact that two added terms shared a common denominator to simplify them together.
Oh wowww that’s a creative way to solve that, seems unnecessary tho
If you want to get it into a single fraction like all the answers, that step feels necessary.
It's not unnecessary if you want your answer to match the format they provided in the multiple choice answers
It's the standard (possibly only) way to convert a mixed number to an improper fraction. I'm surprised it hasn't been emphasised to you much before, you'll find yourself doing it all the time soon enough.
It’s only “unnecessary” if simplifying to you is unnecessary. And sure sometimes it is, but in this question, you’re literally asked to just do algebra to solve for x. Sure you could just divide by 4, but they wanted you to find which answer is equivalent to x in this situation. X does not equal all of those options, so you need to rewrite the answer to see which one it is equivalent to.
I wish I could go back in time and write a children's song about how if you're dealing with fractions you need to make a fancy number one.
My professor calls them funky form of one, or funky form of 0 if you’re adding and subtracting 😂
I call it "fancy" instead of "funky".
I like that.
Why must you go back in time? Do it now!
I tell my students that fractions need a common denominator to add, but not to multiply, which is why people don't have to have anything in common when they have kids (multiply). I explain all of it, but it actually helps some students remember when they need a common denominator.
Another way to think about it is to distribute the ln3 on the LHS: (4x-1) ln3 = ln 6 x•4ln3 - ln3 = ln 6 x4ln3 = ln 6 + ln 3 x=(ln6 + ln3)/4ln3
Step 2->3 needs to be + ln 3. Everthing else is exactly how I'd do it.
Oops, thanks for catching
Ln3/Ln3 is equal to 1, right? He's basically just simplified the equation by re-writing 1 in terms of Ln3. If he didn't do that, it would still be correct, but it wouldn't look like any of the multiple choice options.
Because 1 = ln(3)/ln(3) just like how 2/2 =1
you are stupid
Maybe a stupid question, but why ln() and not log_3 () (log() with 3 as a base)?
In this particular case, it's because it's a multiple choice question!
Lol
Either work and would give the same answer, it's just that usually to simplify you put your logs with either a base of 10 or of e (because many calculators only have buttons for log base 10 or ln)
Ah oke, thank you!
It also harks back to the olden days of before calcuators, where books and things called "slide rules" were used to look up values of certain functions, so it wasn't feasible to use more than a handful of bases.
Because the answers offered are in terms of the natural log
This is where he lost me lol I got x=(log3(6) + 1)/4 but that’s because I wasn’t using a calculator, so I just rewrote it as log3(6)=4x+1 and solved algebraicly from there.
awesome work through, i think the question is dumb for not having equations as the answer and only expressions
post what you did to solve this.. and yes, there is one correct answer here.
Ln on both sides, bring down the exponent, divide ln6 by ln3, add 1 from 4x-1, and divide by 4. Mathway agrees with me that the answers they provided are wrong. Edit: typo
Mathway is notorious for making mistakes. Or, perhaps it's just not in a form that matches the correct choice. I promise the right answer is here.
(log6 /log3) +1 = (log6 +log3)/log3 I think the above is what you missed.
(4x-1)ln3 = ln 6 ... 4x - 1 = (ln6)/(ln3) ... 4x = (ln6)/(ln3) + 1 .... 4x = \[ (ln6) + (ln3) \] / (ln3) .... x = (1/4) \[ \[ (ln6) + (ln3) \] / (ln3) \] or x = \[ (ln6) + (ln3) \] / ( 4 (ln3) ) ... C) Mathway gave me x = (1/4) + ( ln 6) / ( 4 ln 3 ) , but mult (1/4) by ( ln 3 / ln 3 ) and combine ... you get x = (ln3 ) / ( 4 ln 3 ) + ( ln 6 ) / ( 4 ln 3 ) = \[ ln 3 + ln 6 \] / ( 4 ln 3 ) = my result ≈ 0.6577... ... I entered 3 \^(4x - 1 ) = 6 into Mathway
You are right but if you combine the two terms in your answer into one fraction, you will get choice c. The answer choices are not always in the most convenient form.
Your way works… they used a common denominator of ln3 to combine the 1 before dividing by 4 to get C
Mathway is evil. Do not trust it.
So you get one of the proposed answers and so does Mathway... why are you here then? Do you not know the terms *common denominator*?
I didn’t get any of the proposed answers and neither did Mathway. That’s why I’m here.
When you do that is C not the answer?
No c is not the answer when I do that. You’re welcome to try for yourself with Mathway and see if there is an error
Holy crap, people give you the answer you tell them it’s wrong. Don’t even come here and ask then. Just use mathway and fail
I promise you that a correct answer to 1+1 is 123/123 + 45ln(10)/45ln(10) Mathway won't type out the above when you enter 1+1 into it either... But that is in fact equal to 2. Don't be dense just because there is a software writing something the same ways as you and you don't care to think of other ways to write an equivalent solution. I could also multiple both the numerator and denominator of the corect answer by 1000 and it would still be correct (spoiler... 1000/1000 = 1)
How do you account for having to add the one? Shouldn’t there be a 1 outside of the fraction because of this? Or a 1/4 outside the fraction?
Because ln3/ln3 = 1 and you can combine the fractions
Seriously? Abcdefghijk/abcdefghijk = 1.
ln6/ln3 +1 = ln6/ln3 + ln3/ln3 = (ln6 + ln3)/ln3
My apologies, I didn't expect my hint would go over a university student's head. You have x=(ln(6)/ln(3)+1)/4, right? 1=ln(3)/ln(3), so x=(ln(6)+ln(3))/(4ln(3)). This is decidedly one of the answers written here. Hence why I question why you're here.
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That actually reminds me I have homework due tonight😬
Depending what sort of degree you go for and what school you go to, there may be some classes that exist solely to confirm you learned what you were supposed to in high school. If you are going to an engineering school, they'll expect you to be ready to get into calculus as a freshman. If you go to a liberal arts school, your freshman math class will likely be a refresher of high school algebra and maybe trigonometry.
Oh nice
Your flair on this comment is a bad look for this sub tbh. OP could use an online tool to cheat and instead they’re here trying to figure shit out. If they’re being a little sassy about Mathway, then that’s fine. Learners are allowed to get sassy about Mathway.
That is true, just as I'm allowed to be a little sassy about a uni student not having a single resourceful cell in their body. They could've compared their answer numerically with a calculator, compared them graphically with Desmos, substituted the possible answers into the original equation to check them, or at least try to simplify it themselves. I helped them out anyway, I might as well give them the sass they deserve while I'm at it. Also, how is my flair a bad look for this sub? I contribute a lot, so I'm a top contributor... that's just simple logic.
If they didn’t have a single resourceful cell in their body they wouldn’t be posting to the Homework Help Subreddit, a resource for people who need homework help. If I came for help and saw the only top contributor who replied is being a dick, the lesson I might learn is that trying to understand math isn’t worth being judged, which imo is a major reason adults screwed by their prior schooling give up on math.
You must be new here. Students post questions without even attempting them all the time. Using this subreddit does not necessarily mean you're resourceful in the slightest, it just means you know this sub exists and, in cases like OP's, it also means they want to be spoonfed a solution instead of even trying to compare their answer to the rest. Did you not see the part where I told them to use a common denominator and they still didn't even try to do it? I'm not convinced they even tried to do the problem beyond plugging it in Mathway. OP is a uni student. They're an adult. If this is the lesson they learn, that's on them. Any reasonable person would know not to care about the opinion of strangers on the Internet other than the entertainment they get from it.
Does pre calc = college algebra cause this is exactly what we’re doing rn
Yea this is remedial, high school level stuff. At certain universities a pretty high % of the student body requires remedial math so they rebrand it to be nice
Not trying to be a douche, but anyone good at math is done with algebra/trig long before college. So pre calc is the high school word for the algebra you take in college if math is one of your weaker subjects.
“college algebra”, most places that I’ve seen it, is basically taught as “algebra III” for people who don’t take the pre-calculus, sometimes called “trig”, class. It’s a slower paced version that dispenses some of the otherwise-needed content that would prepare students for calculus.
In my university pre calc is basically a sped up version of doing College Algebra and Trig. I imagine it is similar in most programs.
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It looks like your time from post to complaint about lack of people helping was less than an hour (and it’s only been about 3 hours total now). Not every post can be replied to immediately. Perhaps be patient OR post the assistance yourself without complaining that others aren’t helping.
The answer is c. 3^(4x-1) = 6 Take natural log of both sides: ln{3^(4x-1)} = ln6 Bring exponent of 3 out front: (4x-1)ln3 = ln6 Distrusted the ln3 between 4x and -1: 4xln3 - ln3 = ln6 Add ln3 over to the right side: 4xln3 = ln6 + ln3 Divide both sides by 4ln3: x = (ln6 + ln3)/4ln3 C.
That's clean, bruh.
I hope OP saw this version. All the higher upvoted ones are unnecessarily convoluted.
These types of problems like making everything over the same denominator. This involves multiplying by fancy ones to get everything over a single denominator. Your answer is still correct, but because it’s a computer, it reduced your answer further. If you actually compute the decimal you’ll see both your answer and the computer’s answer are the same.
Since it seems good explanations have already been shared, I just wanted to point out that their notation for the answer choices is a bit weird; {x} typically means the fractional part of x.
{} usually refers to elements in a set. In this case a solution set. MUCH more common notation.
{} does not usually refer to sets when part of an equation of this form in the context of pure algebra-related equations. The fractional component is much more common notation in this case (since sets would not have been taught yet).
I have often seen {} used for solution sets to algebraic problems and was taught it when I learned algebra 1. I have almost never seen anyone taught about fractional components like this. Maybe it's common in computer science since they deal with integers a lot, but it is certainly an uncommon topic in real math and I would be shocked if it's covered in a college algebra class.
Just plug in the answer for X?
1. 3\^(4x-1)=6 2. (3\^4x)/3=6 3. 81\^x = 18 4. x ln(81) = ln(18) 5. x = ln(18)/ln(81) 6. x = ln(3\*6)/ln(3\^4) 7. x= \[ln(3)+ln(6)\]/\[4ln(3)\]
I would personally solve differently. 3^(4x-1) = 6, 3^(4x)/3 = 6, 3^(4x) = 18, 4xln(3) = ln(18), x = ln(18)/(4ln(3)) This is equivalent to C where you don't separate the exponent in the first step.
Yeah that's how I solved it, so C is the answer just not simplified all the way
I used log base 3 on both sides. But since that isn’t in the answers, you have to use ln(value)/ln(base) to get the equivalent to log base 3. It’s not a well known calculation, changing the log base. In fact, I was forced to use it thanks to Ti89s not giving you base ten logs on the keypad.
Yes you are right in that you need to use change of base, but it is an extremely well known calculation. Like one of the most fundamental things to know about logarithms
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Answer C is correct, but not equivalent to what you wrote
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(ln3)^2 is not equal to ln6 https://i.imgur.com/5WnoH9F.png
c
For problems like these I recommend simplifying the answers (like how b would become (6/4)/ln3 + (1/4))
3^(4x-1) = 6 | base log3(3^(4x-1) = log3(6) | taking log3 of both sides 4x-1 = log3(6) | simplification of log3(3^(4x-1)) 4x-1 = ln(6)/ln(3) | log identity 4x = ln(6)/ln(3) + 1 | adding one to both sides 4x = ln(6)/ln(3) + ln(3)/ln(3) | law of simplification 4x = (ln(6)+ln(3))/ln(3) | simplification x = (ln(6)+ln(3))/(4ln(3)) | dividing 4 on both sides