solving a logarithmic equation with different bases
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- เผยแพร่เมื่อ 15 ก.ย. 2024
- How do we solve a log equation with different bases? Here we will see how we can use the change of base formula for logarithm to solve log_4(x)+log_2(x)=6.
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Could you just have done:
ln(x) + 2 ln(x) = 12 ln(2)
3 ln(x) = 12 ln(2)
ln(x) = 4 ln(2) = ln(2^4) = ln(16)
x = 16
Too simple is not chinese ...
Even don't need ln
log2 x + log4 x = 3log4 x = 6
log4 x = 2
x = 16
That’s what I was thinking, just add the ln(x)’s and then divide both sides and it’s simple from there
@@bitterberry_ This particular example works out nicely like that due to the bases being 2 and 4. But I think the point is to show a more general method
@@BigDBrian You should always use the simplest approach for the task at hand!
Why not recognize 4 is 2^2 and change log4(x) to log2(x)/log2(4) and continue from there? Much simpler.
Yeah, I also think it makes more sense to be a little bit more picky choosing the log base when you can choose any base you want. Better think what is the simplest one to use instead of sticking to natural logarithm just because it's the one you're most used to. I always say it's better to think a little instead of following formulas mechanically. And personally I favour base 2 logarithms anyway, because they're the ones you'll find most commonly in computer science. :)
Mathmaticians never seek for easy way
Or just use that property when power of base goes in denominator 🙂
Theres also the short cut formula of:
Log x^n(y)=Logx(y)/n
@@nemanjalazarevic9249 yes, it is a basic log property
Just a thought - If you use a base change to "2" instead of "e", you rapidly get to (log2(x))/(log2(4)) + log2(x)=6; 3/2*log2(x)=6 or log2(x)=4, hence x=2^4
You have the "best" way to do it
Love ya
Where u get 3/2
@@cipherbenchmarks : Factor out the log2(x)) from (log2(x))/(log2(4)) + log2(x)
(log2(x))(1/(log2(4)) + 1) and as log2(4)=2, (log2(x))(1/2 + 1)=(3/2)log2(x)
Hope that helps.
@@animalfarm7467 I appreciate you responded after three years and yes, it helped
Why not :
1/2log2X+log2X=6
3log2X=12
log2X=4
X=2^4
X=16
Exactly what I did in my mind when I saw the thumbnail.
+)
donati880, that is also correct.
Exactly,
Yes this is very easy and simple method.
As others have noted, observing that the bases are both powers of 2 suggests that the best base to change to would be 2. That then means the denominators in the change of base become nice integers and the result then trivially simplifies to log2(x) = 4. Of course, your way works more generically and shows off more log properties.
I think base 4 would be even better here
I like this video especially because I can see the answer beforehand. I did it differently, but it uses the same reasoning.
LOG4(x) + LOG2(x) = 6
LOG4(x) + 2(LOG4(x)) = 6
3LOG4(x) = 6
LOG4(x) = 2
x = 4² = 16
You can do that last step a little easier. When you have
ln(x) + 2ln(x) = 12ln(2)
You can simply add the ln(x) on the left since ln(x) acts like a variable and get 3ln(x) = 12ln(2)
Then divide both sides by 3 to get
ln(x) = 4ln(2)
Then put the 4 into the ln(2) to get
ln(x) = ln(2^4)
Then x = 2^4, which is obviously 16
Not bragging, but this was too easy. I literally solved it in my head just looking at the thumbnail.
Your videos are awesome tho.
Same
It's still really cool math so I don't mind
Same
Yeah, same
Easy for us in higher grades. Consider the people who are not familiar to the concepts of logarithms🙃
i prefer:
ln x + 2 ln x = 12 ln 2
3 ln x = 12 ln 2
ln x = 4 ln 2
ln (x) = ln (2^4)
x = 2^4
x = 16
same here, just think it's unnecessary to bring the 2 into ln x
Also that we can convert Log X_4 = 1/2 of Log X_2. So both terms are at Base 2, taking common the Log portion).
3/2* LogX_2 = 6 >>> X^3/2 = 2^6 >>> X= 16.
I did the same thing , is it a right way to solve logarithmic equations that can be converting to be the same bases ??
I rarely can solve math problems on this channel in my head, or even solve without pain but this problem is super simple or I got some wierd inspiration. Answer just poped in my head Sheldon style.
log4(x)+log2(x)=6
log2(sqrt(x))+log2(x)=2 +4
log2(sqrt(x)) = 2 and log2(x)=4
x= 2^4 = 16
Maybe it's time for us to make it *COMPLEX*
BlackPenRedPen : Takes 6 mins to solve the world's easiest log question
Me : Solves it in literally 10 seconds in head
LITERALLY?
@@blackpenredpen According to google
Literally definition : used for emphasizing something
He's tutoring, you're not.
ln x and 2 ln x are like terms
Kevin vs. Gamingz yup! It was 10:30pm when I recorded this..
What is log_i(i!¡) ?? (Logarithm with base i of i-th tetration of i subfactorial
I found another way!
Note: 2*log_4(x)=log_2(x)
log_2(x)+2*log_2(x)=12
log_2(x^3)=12
x=2^(12/3)=16
ln(x) + 2×ln(x) = 3×ln(x)
Instead of doing ln(x)/ ln(4), why not just do log2(x)/log2(4)? Then you get log2(x)/2 + log2(x) = 6.
Then we have (3/2)log2(x) = 6
log2(x) = 4
x = 16
This just seems a lot easier.
Love the videos.
surely i had a problem in solving that,thanks dude
Given:
lq(x) + lb(x) = 6
Knowing lq(x) = ½lb(x); as lb(4) = 2:
½lb(x) + lb(x) = 6
Adding fractions:
(3/2)·lb(x) = 6
Multiplying both sides by ⅔:
lb(x) = 4
Exponentiating both sides using 2 as the base:
x = 2⁴
x = 16.
Here, lq(x) = log(x) with base 4, lb(x) = log(x) with base 2.
At 3:18 I was so stunned about what you did next that I had to check the calendar to make sure it wan't April 1st...
Haha, it could be written just 3ln(x).
Around 3:30, you could just do ln(x) + 2ln(x) = 3 ln(x) and go from there, quicker and simplier
I have just applied 2 to base of logarithm and saw that (log2^x/2) + (log2^x/1) = 6 And then i have multipled second term by 2 then saw that (log2^x/2) + (2log2^x/2) = 6 this turned into 3log2^x/2 = 6 and did the equation which turned into log2^x = 4 and simply found the x equals to 16. I have watched so much of your videos. And w/o you i couldn't be able to find out the answer. All thanks to you
I really enjoyed how you showed how easy to change the base. Great!
Why not add ln(x)and 2ln(x) to get 3ln(x) then divide both sides by 3? Much simpler than the route to the solution shown in the video (IMO).
I agree, but it has the advantage of showing more rules on logarithms
Remember he teaches stuff.. He wanna use as many different methods he can, without confusing people :>
I agree, Delta & Rune, that he probably did it the way he did to include more log laws. But I think taking the simpler route here would have been better.
Those other rules could have been shown in another problem (or video). ;-)
Thad Spreg
Honestly... I totally forgot about it. It was 10:30pm when I recorded this at school after 5 hours of teaching...
blackpenredpen you work very hard. We appreciate all your efforts!
At 3:30 just simplify to 3ln(x)=12ln(2) and divide both sides by 3 to get ln(x)=4ln(2), simplify to get ln(x)=ln(16), cancel the 'ln's and you get x=16
To avoid confusion I would just solve log(x)(1/log4 + 1/log2) = 6 for log(x) and then
set x = 10^log(x) = 16, where log is to base 10.
I did it like this.
Let y=log4 (x)
Let z=log2 (x)
y+z=6
4^y=x
2^2y=x
2^z=x
2^2y=2^z
2y=z
y+2y=6
3y=6
y=2
log4 x = 2
x=4²
x=16
Nice.
Easier method,
log_4x + log_2〖x 〗=6 log_(bxb…..ntimes)〖a 〗=1/n( log_ba )
〖1/2 log〗_2x + log_2x =6
3/2 log_2x =6
Dividing and multiplying both sides with 3 and 2 respectively
log_2〖x 〗= 4 log_ba=x then a=b^x
Therefore, x=2^4 =16
A “hidden” rule I found with logs is that if you have log_b(a), it will be equal to log_(b^2)(a^2) or alternatively 2log_(b^2)(a).
This is because if you use your change in base formula and exponentiate the arguments to the same number, then you can use the power rule to take the exponents out of the argument and cancel, of which would leave you with another fraction to apply the change in base formula.
Example:
log_9(4)
ln(4)
---
ln(9)
ln(2^2)
---
ln(3^2)
2ln(2)
---
2ln(3)
Simplify to
ln(2)
---
ln(3)
And
log_3(2)
Hence, log_9(4) = log_3(2)
Correct but there’s still a simpler way to solve it
log2 (x^0.5) + log2 (x) =6
log2 (x^1.5) = 6, we know log2 (2) =1
Let introduce log2 in the second member
log2 (x^3/2) = 6log2 (2), (we know 3/2 = 1.5)
Let’s write 6log2 (2) in the form of exponential
log2 (x^3/2) = log2 (2^6)
Then the log2 canceled out
X^1.5 = 2^6
X= 2^6/1.5
X= 2^4
X= 16
Not really a math question, but how do you write on a whiteboard using two pens? I kind of wanna see a close up lol.
Easy after a little tutelage in legerdemain. Manipulating multiple objects in one hand.
Fred
Logarithmically Transform the function y=(〖(3x^2+2)〗^2 √(6x+2))/(x^3+1)
Log2(x)=2log4(x) so,
3log4(x)=6
Log4(x)=2
By definition,x=16
log₄x + log₂x = 6
Well, because 4 = 2², we can write
log₂x = 2·log₄x
And then we have
log₄x + log₂x = 3·log₄x = 6
log₄x = 2
x = 4² = 16
Note that the exponent, 2, in 4 = 2² is the log base 2 of 4: log₂4 = 2.
So in more general cases, when there are logs of more than one base, they can all be converted to a common base, using
logᵤx = logᵤv · logᵥx
You can choose any base you want as the common base - whatever works best.
Now to watch Mr. Pen, to see how he does this . . .
Ooh, yes that's the right answer, but sorry to have to say I don't think this is the best way to get there. Once you have:
lnx + 2·lnx = 12·ln2
just combine the LHS to get:
3·lnx = 12·ln2
lnx = 4·ln2
x = 2⁴ = 16
Kudos for giving the change-of-base formula - that's a powerful tool in log problems!
Fred
another solution :
Suppose that log4(x) = y
If 4 ^ y = x - - or 2 ^ 2y = x
We take log2 (x) in the last relationship
If 2y = log2 (x) then we substitute the original equation to find y
y = 2 After compensation, we find the value x
x = 16
3:25 alternatively,
lnx + 2lnx = 12ln2
=> 3lnx = 12ln2
=> lnx = 4ln2
=> x = e^(ln2*4)
=> x = 2^4
=> x = 16
You could also have taken log 2 getting log 2x +2logx=12 then log 2 x • x^2 so then log2 x^3 and then proceed the same way
I know we like to use the natural log for the general case, but for this specific case, wouldn't it have been simpler to convert the log_4 to log_2?Then you would have had (1/2)log_2(x) + log_2(x)
I use it and I get the solution X=8*(cubicroot of 2) "its soo close to 16" and I get the same solution when I use the Mathway website .
Am I get the right solution ??
when you got lnX + 2lnX = 12ln2
you should just add the lefthand side, then divide everything by 3.
So you'll get to lnX = 4ln2 = ln16 a lot faster and don't have to take the cube root which creates extraneous solutions
So this equation contains 3 solutions,since x^3=2^12?
No, this equation is not a cubic equation, where are you getting the idea that this equation is a cubic equation? (3/2)log2(x) = 6 => log2(x) = 4 => x = 2^4 = 16. Turning it into a cubic is not only unnecessary, but it changes the equation itself, and therefore it’s solution set. You are not allowed to simply create irreversible functions in a manipulation.
Dear Jenna Grace, log[(4x)(2x)]=6. Now this: log(8x^2) = 6, a base 10 logarithm. Change it to exponential 8x^2=10^6 ; x=sqrt of 10^6/8
You can use the property where log x to the base b^n is equal to 1/n times log x to the base b.
name of that property?
here's how i solve it: 4_x means log base 4 of x, so we have the identity 4_x = 1/x_4 = 1/2 x_2 = 2_x /2
so 4_x + 2_x = 3/2 of 2_x = 6, we get 2_x = 4, x=16
Given that log_a p=0.7and log_a q=2, find log_a〖p^2 〗, log_a〖p^2 〗 q and log_a (apq)
Sir you could directly write..
Log[4]x=(1/2)log[2]
HERE I USED SQUARE BRACKETS [. ] TO DENOTE THE BASE OF THAT LOG
This could become more easy
I think this method is also possible!
Log 4 (x) + Log 2 (x)= 6
(Log 2 (x))/ Log 2 (4) + Log 2 (x) = 6
Let t = Log 2 (x)
t/ Log 2 (4) + t = 6
t/ 2 + t = 6
3t/2 = 6
t= 4
Log 2 (x)= 4
x = 2^4
x= 16
I just went by the definition of the logarithm: 4^2 =16 = 2^4. So if x=16 then log_4(x)=2 and log_2(x)=4. The sum is 6. This, of course, worked only because x is an integer and the logarithms as well.
It took me 5 steps to finish this and i feel like he makes it complicated, i mean why don't you do like this for quicker?
Step 1: we can turn the question into
1/2log2(x) + log2(x)=6
Step2: log2(x)(1/2+1)=6
Step3: 3/2 log2(x)=6
Step4: log2(x)=4
Step5: x=16. DONE!
Another option:
3lnx = 12ln2 ; divide by 3; rewrite RHS as ln2^4 and then antilog...etc.
Another way I found to solve this:
From lnx/2ln2 + lnx/ln2 = 6
Factor out (lnx/ln2):
(lnx/ln2)(1/2 + 1) = 6
(lnx/ln2)(1.5) = 6
lnx/ln2 = 4
lnx = 4ln2
x = 2^4 = 16
How about solving a exponential equation with different base?
Like this
2*5^x-7^x=1
I have tried for at least 2 hour but couldn't find the solution yet.
If you help me find its solution please!
I'm high-schooler so i don't know if this is the right way for solving but here is my solution: hizliresim.com/oXd2rk
I think there can be a mistake when we using the root because we find x=0 but root is 5^x it's equal to 1 :P
havent tried but you might need to use the lambert w function
Jensen No, that is not sufficient. Equations of the form a^x + b^x = c cannot be solved in the general case analytically, not even using the W function you mentioned. You can only solve it if a = b or if c = 0 or if ab = 0 or if a = 1 or if b = 1. To visualize how this cannot be solved in the general case, all you must show is that this equation requires solving a pseudo-polynomial, which also cannot be done.
@@angelmendez-rivera351 Could you link me to somewhere where i can find more about what a pseudo polynomial is? i can only find computing pesudo polynomial runtime algorithms
Jensen Well, I used the term “pseudo-polynomial” a little differently. Notice that a^x = e^[ln(a) x] and b^x = e^[ln(b) x]. So a^x + b^x = e^[x ln(a)] + e^[x ln(b)] = [e^x]^[ln(a)] + [e^x]^[ln(b)] = c. Let y = e^x, so the equation is now y^ln(a) + y^ln(b) = c. I called it pseudo polynomial because it is similar to a polynomial equation, but it is different in that the powers are not from integer exponents, unlike true polynomials. But these equations cannot be solved analytically. Something so simple as y^ln(2) + y^2 = c cannot be solved without numerical approximation. Not even with W can we do anything about it.
So perfect so beautiful i can't believe it. Never seen something so cool in maths with such hard functions like this.
You are a very good teacher.
YOU ARE STILL SAVING LIFES
ln(x)+2ln(x)=12ln(2)
3ln(x)=12ln(2)
ln(x)=4ln(2)
ln(x)=ln(2^4)
x=2^4
x=16
It’s mathematically the same, it just feels better doing it this way for me idk
you can either apply change of base law with ln or log_2, OR you can do the other way.
log_4(x)+log_2(x)=6 (1)
(1) --> log_4(x) = 6-log_2(x)
let f(x) be log_4(x) and g(x) be 6-log_2(x)
f is constantly increasing while g is constantly decreasing, thus there can be only one solution.
By inspection, x=16 is a solution, therefore, 16 is the only solution
**P.S. you could let f(x) = log_4(x)+log_2(x) and g(x) = 6 and stick to the same method, you will see that f is constantly increasing while g is a constant function, thus there can only be one intersection point at the most.
Please integrate tan inverse X whole upon 1 + x square to the whole power 3 by 2
I think this is easier to use the change of base formula on just the first addend only using log(base 2). This gives
log(base 2)x / 2 + log(base 2)x = 6 → (3/2) log(base 2)x = 6 → log(2)x = 4 → x = 2⁴ = 16.
Just have log2(x) as "t" and then t+t/2 = 6 which gives t= 4 so log2(x) = 4 or X= 16.
Simple method here the easiest....log 2^2^x+ log 2^x=6
1/2 log 2^x + log 2^x =6
Let log 2^x be k
1/2 k+ k =6
3/2k =6
k=4
But log 2^x =k which 4
Log 2 ^x=4
2^4=x
X=16
Simplest way and easiest
One of my favorite bprp videos♥️
log(x)/2log2 + logx/log2 = 3logx/2log2 = 6, 3logx =12log2, logx/4log2=1 , logbase16(X)=1 so that means x =16
Easier to keep the coefficients outside of the ln. 3lnx = 12ln2 then x=2^4.
but
log(a^x)(b^y) = (y/x)log(b)
so you could like
log(4)(x) + log(2)(x) = 6
0.5log(2)(x) + log(2)(x) = 6
1.5log(2)(x) = 6
log(2)(x) = 4
x = 2^4
It’s crazy how many math problems I do and I remember doing hundreds of these just months ago and I never retain the rules
I see that spiderman is a patron XD
Good example for understanding the logarithms' law of change of base
This one was easy. Could also use substitution to simplify things even further after change of base
Just notice log4 is always a half of log2 and it can be rewritten as log4+2log4=6 which then can be written as 3log4=6-> log4x=2 x=4 to the power of 2=16
Everybody else saying there's a simpler way. My issue was that there was an unnecessary JAZZ PIANO playing.
*classical piano
@@itscalledatilda There is indeed classical piano in the first 23 seconds. I think I skipped that last time. I was talking about the piano from 4:00 onwards, which I assumed he added deliberately. Looks like he was just next to a music room. Interesting that this problem uses 2^12, given the importance of 2^(1/12) in music (it's the frequency ratio of 2 notes a semitone apart.)
Could you do a video on the logarithmic integral and how to solve it? Everytime I try to solve it on paper it's weird and when I try to use a calculator and integrate the integrand I get the logarithmic integral again
Here is a good one: can one use the Lambert W function along with all the other elementary functions we are acquainted with to solve an equation of the general form e^x = Ax^2 + Bx + C ? And if so, what is the general solution in terms of A, B, and C?
At second step if you just set the equation as
log2(x)/log2(4) + log2(x)/log2(2) = 6
It would be much easier since log2(4)=2 and log2(2)=1
Log(3^x-1. + 3²-x)=Log7 +Log4/log2
Those values and variable in the bracket are all in exponent
I need help
Very good exercise. Thank you!
We could also resolve the first term into base 2 it's easier that way
Wow, I'm seeing a lot of different ways to solve this. Let me throw my 2 cents in (btw, if you do it this way, you'll lose points on most exam)
See both are powers of 2
Try 8 as an input, see that it's short
Try 16, it worked.
TBH I knew it was 16 without doing all the math... But I did it more by converting it back to Exponents... 4^y=x and 2^z=x Y+Z=6 From there 4^y=2^z Bring up a power to make it base 2 2^2Y=2^Z... 2y=z and Y+Z=6 from there 3Y=6 Y has to be 2 and Z 4 from there plug in to get X=16
You are the best professor in life
Thank you
I'm quite new to logs but realised that log_x(y) is the same as log_(x^n)(y^n) so you can rewrite log_4(x) as log_2(x^1/2) and proceed to solve log_2(x * sqrt/x) =6
Oh my god I figured out a bprp problem before he finished on a three year old video about algebra 2 LETS GOOOO
I recognized that since we have logs of 4 and 2 which can both be written as a number base 2 I let x = 2^m. Then you get m/2 + m = 6 or m = 4 so x=16 :P
Isnt there also 16e^i*2pi/3 and its conjugate if we talk complex?
These videos are the only revision I do
Could you demonstrate the property you just used?
(The one to make any log the base that you want)
I woulda just combined the terms on the left to 3*ln(x), then divided both sides by 3, yielding:
ln(x)= 4*ln(2)
Incredible work
Is there way to know a faction by using pictur of this faction?
I really liked the conclusion. (That must make you feel real smart.)
Can you make a video on modular arithmetic?
Factor lnx at step 2 and change to expo. form. Fewer steps gollow
Amazing work man
in 4th line we can also write: ln x + 2 ln x = ln x (1 + 2) = 3 ln x = ln x ^ 3
I love this vidéo it s still good.
Pls, how can i solve log2(8)=log5(x) ?? I really can't figure this out, I need help as fast as possible.. Thanks in advance
Not entirely sure where the difficulty in this problem stems from. First note the LHS is equal to 3 since 2^3 = 8 then from the definition of the log, 5^3 = x so x is 125.
You chose the most complicated way to solve this.
Coulda used difference of cubes. (x-16)(x^2+16x+256)=0
4:28 why do you say "in the real world", you mean you don't use imaginary and complex numbers here or what ?
Can you please help me with this number as well, log125/5_ log16/2+log9/3 .(5,2 and 3) are the bases
What about a more general equation log_a(x) + log_b(x) = c
ln(x)/ln(a) + ln(x)/ln(b) = c
ln(x)(ln a + ln b) = c(ln a)(ln b)
x = exp((c * ln a * ln b) / ln ab)
can we simplify that further? not entirely sure we can, as product of logs doesn't really mean anything
so in this case, a = 2, b = 4, c = 6, x = exp (6 * ln 2 * ln 4 / ln 8) exp (6 * ln 2 * 2 ln 2 / 3 ln 2)= exp (4 ln 2) = 2^4 = 16
If the bases "line up" i.e. one is an easy power of the other, you could use x = a^((c log_a(b)) / (1 + log_a(b))) instead. Here log_a(b) = 2 so x = 2^(6 * 2 / (1 + 2)) = 2^4 = 16