It's the opposite of exponenciation, your x in a logarithm is the exponent you need to give to the base (10 in this case) so that the result is equal to 80 (which is called the argument I think). Not that hard to grasp as a concept
You could also solve this problem using logarithmic properties: 1. We take the log() of both sides, which allows us to put the exponent as a factor to the logarithm of the base. Giving us: "(x-1)log(40)=(2x+1)log(2)" 2. We multiply into the parenthesis on both sides, and isolate the terms with x in them. Giving us: "x log(40)-2x log(2)=log(2)+log(40)" 3. We isolate x by factoring, and dividing the resulting parenthesis to the right. Giving us: "x = (log(2)+log(40)) / (log(40)-2log(2)) 4. Due to the rule of logarithms used in (1), we can express the "2log(2)" as "log(2^2)" which equals "log(4)", whereafter we use the rule of sums and differences of logarithms. Giving us: "x=log(80)/log(10)" 5. Due to log-base10 of 10 being 1, this gives us the result as: "x=log(80)"
Before even watching the video I know I'm gonna hear "Lets put a box around it" and "How exciting". Almost forgot the “Nice” every time 69 or 420 shows up. And you can’t forget the “this looks like a fun problem” when the video starts.
i did it by applying log to both sides, and then solving. it makes it easier because after that its just manipulation. but you do need the value for log 2 and log 40.
I’m having flashbacks to helping my daughter with Algebra 2 😮 I love how you manage to keep the presentation so brisk, yet also don’t have any mystery steps.
My math teacher in high school wouldn't have allowed log 80 as an answer because it can be reduced like square root of 80 can be reduced. Because it is base 10, log 80 = log 10 + log 8 since log ab = log a + log b if the base is the same. Log 10 = 1, so log 80 = 1+ log 8.
Depends, sometimes it’s the opposite. With a class that I was in, they would want me to write 1 + log 8 as log 80. But log 8 on its own would need to be simplified to 3log 2.
A video about a list of the usual "tricks" that can be used in such math problems (like the 30-60-90 triangle, the exponent substitutions you used here, etc...) would be amazing! Thanks!
This is the first time I watch your videos and honestly they're actually helping me understand math even better, I don't know, I try to resolve the problem a little faster than you just to see if I can do it, sometimes I get it wrong but it's fine, I keep trying, thank you! you're helping someone get better at math!
I tried solving it before the video, ended up with the same answer. I just did the log base 2 of 40 first to get the exponent bases to 2. got something around 5.32. set the exponents = to each other due to same bases and boom basic algebra to get to the same decimal answer. (with calculator ofc) nice video tho man 👍
For the people trying to show off your logarithmic skills, Andy intended this problem to be understood by those who aren't really familiar with log. So, instead of using log identities from the get-go, he used log only when the equation was already simplified. You and I may be familiar with log, not everyone is.
Easy method here Just take log on both sides and multiply it in side then you should have an equation with both sides having x of power one and substitute values of log40 ( log4 +log10) and log 2 Then u get 1.602x-1.602=0.601x+0.301 On bringing x oneside and constant on other side u get 1.001x=1.903 Which is approx answer x=1.903 If you need more precise then increase the decimal numbers That's it not even a 2min task
That's what i was thinking it took me about 1 min to solve this mentally i was like he must have asked a numerical value i solved log 8 ans 3log2 = 3X.30 and i got 1.90
2:31 I think explaining that with the inverse function is more logical. Just like in _tan x = 2_ we take the inverse function of _tan_ on both sides, which is _tan^-1_ that gives us _tan^-1 (tan x) = tan^-1 (2)_ which simplifies to _x = tan^-1 (2)_ Here, the inverse function of _10^x_ is _log 10 (x)_ Taking that on both sides gives us _log 10 (10^x) = log 10 (80)_ which simplifies to _x = log 10 (80)_
Hi man! What an amazing video! But, the verification can be done like this without a calculator: 40^(x - 1) = 40^(log80 - 1) = 40^(log8 + log10 - 1) = 40^(log8 + 1 - 1) = 40^log8 2^(2x + 1) = 2^(2log80 + 1) = 2^(2(log8 + log10) + 1) = 2^(2(log8 + 1) + 1) = 2^(2log8 + 2 + 1) = 2^(2log8 + 3) = 2^2log8 * 2^3 = 4^log8 * 8 = 4^log8 * 10^log8 = (4*10)^log8 = 40^(log8) here, we can see that 40^(x - 1) gave us 40^log8 and 2^(2x + 1) also gave us 40^log8. so, verified.
@@rathersane You can use logarithm regardless but yes, it's usually 10 as base. In this case, it works better. I'm just saying what I prefer in my opinion. Is that bad?
So, just for the sake of it, 40 is equal to 2^5.321928. If I understand powers right, this would give: (2^5.321928)^x-1 = 2^(2x+1) 2^(5.321928x - 5.321928) = 2^(2x+1) 5.321928x - 5.321928 = 2x+1 3.321928x - 5.321928 = 1 3.321928x = 6.321928 x = 6.321928/3.321928 x = 1.90309
I knew that because I brute forced and put 1 as x and 2 as x and then just did 1.5 and kept raising it to 1.9.... They need to make it a bigger x so that I can't just input numbers quicker than it takes to actually use the correct formulas😅
just introduce logs to both sides in the beginning 40^x-1=2^2x+1 log 40^x-1= log 2^2x+1 power law of logs, the power of the log can become the coefficient of the log (x-1) log 40 = (2x+1) log 2 x log40 - log40 = 2x log2 + log2 move the x terms to the left, and the log40 to the right x log40 - 2x log2 = log 2 + log 40 factorise x from the right x(log 40 -2log2) = log2 + log40 x= (log2 + log40) / (log40 -2log2) = 1.903089987
There was a really easier way, you could write as (2^ 5.32192809490)^x-1=2^2x+1 ,since both bases are the same we can write down as 5.3219280949(x-1)=2x+1 which is then equal to 5.3219280949x-2x= 5.3219280949+1 then x= 6.3219280949/3.3219280949
(x-1 root 2^(2x-1)) = (2^(2x-1))^(1/(x-1)) = 2^((2x-1)/(x-1)) = 2^((2x-2+1)/(x-1)) = 2^((2x-2)/(x-1)+1/(x-1)) = 2^(2+1/(x-1)) = 2^2 × 2^(1/(x-1)) = 4 × (x-1 root 2) You could move more stuff around to get the x-1 on the top, but it's way easier to just use logs
so simple you just know about log 5on2 =2.3 and done and solve it but you can do it simple you should change this log to log5 /log2 => 0.7/0.3 =2.3 (rounded ) and you dont wana calculator or somthing this question was standard for student
How i would solve: 40^(x-1)=2^(2x+1) log2(40^(x-1))=log2(2^(2x+1)) (x-1)log2(40)=2x+1 log2(40)x-log2(40)=2x+1 (log2(40)-2)x=(log2(40)+1) x=(log2(40)+1)/(log2(40)-2) =(3+log2(5)+1)/(3+log2(5)-2) =(log2(5)+4)/(log2(5)+1) ≈1.9
gazylion pingu 3 razy zabija random sekunde przed wzieciem pojazdu ideal ost sek nie zdazam czegos przez lagi 100 razy pierwszysm strzalem podrzucam czolg drugi pod przelatuje