A Log Equation | log_2n (72)=log_n (18)

[@eckhardfriauf]

2nd method is awesome! Makes it so easy!

[@SyberMath]

Glad you think so!

[@moeberry8226]

The domain restriction on n is not only positive real numbers and not 1. But also n cannot equal 1/2. Since 2n would equal 1. If n=1/2. So n>0 and not equal to 1 or 1/2.

[@vladimirkaplun5774]

Who cares about the domain. Once you receive a root you just check if it fits in.

[@angelmendez-rivera351]

@@vladimirkaplun5774 People who care about correctness care about the domain 😊

[@vladimirkaplun5774]

@@angelmendez-rivera351 once you test the potential answers you may not mention the domain.

[@SyberMath]

Correct!

[@broEye1]

I went more directly. Since log(x,y)=z means x^z=y, x^log(x,y)=y. So, I just applied the equation directly: (2n)^log(n,18)=72, then reduced that to 2^log(n,18)*18=72 2^log(n,18)=4 2^log(n,18)=2^2 log(n,18)=2 n^2=18 n=sqrt(18) n=3sqrt(2)

[@broytingaravsol]

n=​3√2

[@mputuchimezie7966]

You always reserve the best method for last 👏

[@SyberMath]

😊 thanks

[@waiphyoemg1668]

I usually try with the second method when I see such kind of log problems. Btw thank you Sir!

[@SyberMath]

Happy to help

[@goldfing5898]

Set t = log_2n(72) = log_n(18). Now you can rewrite the first equation to (2n)^t = 72 2^t * n^t = 72 and the second equation to n^t = 18. Plugging this into the first equation yields 2^t * 18 = 72. Divide by 18: 2^t = 4. Take the log_2() of this: t = 2. Plug t = 2 into the equation n^t = 18: n^2 = 18 n = +- sqrt(18) = +- sqrt(9*2) = +- 3*sqrt(2) Since the bases (n and 2n) must be positive, only n = 3*sqrt(2) is a correct solution. Ha, I just saw that I chose the 2nd solution :-)

[@infinite1.0]

From seeing that one of the bases is twice the other and the exponent must be the same since they share exponent, then you can see that 72=18*4 And you have: n^a = 18 (2n)^a = 72 Multiply the first one by four and make them equal (2n)^a = 4*n^a Divide n^a from both sides and apply log base 2 after 2 = a Knowing this you can replace in the original n^2 = 18 n = 18 ^ (1/2)

[@SyberMath]

Nice

[@RAG981]

Method 1 is from the masochism school of mathematics. I went straight for your method 2 because that is sane.

[@SyberMath]

😁

[@MichaelJamesActually]

Me forgetting the negative sq rt for the millionth time, and it's finally ok!

[@SyberMath]

😁

[@Ssilki_V_Profile]

Furst method - you forgot to note that N is neither equal to 1 nor equal to 1/2

[@Taxonomyonlinemathematics]

Impressive

[@HariprasadmHariprasadm]

Sir we r with you

[@SyberMath]

Thank you! 🤩

[@-wx-78-]

Change base to e is so natural. 😉

[@sergeyshchelkunov5762]

2nd method is so much faster.

[@kianmath71]

n = e^(ln(18))/0.5 = 3sqrt2

[@arthurnetto3792]

why it doesnt work if I make log72 (base 2n) = [log72 (base n)]/[log2n (base n)] this is equal log18 (base n), cross multiply and we get log2n (base n) = [log72 (base n)]/[log18 (base n)] and the right hand side is equal log72 (base 18) and this is equal log2n (base n)

[@spelunkerd]

All of what you wrote seems accurate. Where do you want to go from there?

[@ramraman5233]

👌

[@tambuwalmathsclass]

First method for the critical thinkers, second method for those that love simple things.

[@SyberMath]

Good point!

[@abu-karz]

Dude literally spawns the idea to come up with a solution for these math questions. I envy this quality.

[@SyberMath]

Hehe. Thanks! 💕

[@angelamusiema]

Si annullano il logaritmi anche se c'è n2 nel primo elemento perché se n e nullo perché anche log2n è 0 ..... ; e si fa 72-18=54 per la semplificazione delle proprietà logaritmiche . A parte che sin dall'inizio secondo me un equazione impossibile.

[@barakathaider6333]

👍

[@AdityaKumar-gv4dj]

n=3√2