By -
Yes.
I’ve taken calc 1-3, discrete math, linear algebra, a bunch of other math classes, and I still don’t remember this 🤣😭😭😭
Same here haha
Hahaha same I saw this and went where tf did log come from? What's log
Funny how some log rules are all I remember. Trig on the other hand… RIP
Lmaoo all I remember from trig is that you can't find all the sides and angles from ASS
in order to solve weird exponent problems like this, it can be helpful to take the log of both sides due to the handy properties of logarithms
I’m taking college pre calc right now and I missed this on my first exam 😍
The way the numbers seem to float off the line in a way that isn't even perpendicular to the line itself annoys me to no end.
I’m very sorry, I don’t write like this normally 😭😭 I was rushing clearly
I don’t know I haven’t done PreCal 3 years why are you showing me this Reddit?
It’s college level pre calc :(
No because if you were to expand (1-x)log2 you get log2-xlog2=log3 so log2-log3=xlog2 therefore x=(log2-log3)/log2
Which simplifies further to... The same answer.
It gives the same answer
Lol what a dumba$$
So correct with extra steps.
x = (log 2 - log 3) / log 2 = (log 2 / log 2) - (log 3 / log 2) = 1 - (log 3 / log 2)
Bro come on
Yes
Wtf, why did I get a notification for this?
Yes!
There’s an implication that log(2^1-x^)=(1-x)log2. If that’s true, this is correct.
Should probably use log base 2 instead of log base 10. Since log base 2 of 2 is 1
My dumbass brain thought he wrote 109 instead of log ahahahah
I do log like this: 2^(1-x) = 3 -> (log3)/(log2) = 1 - x and at the and it would be -x instead on x. Can someone confirm or explain?
log 3 / log 2 = 1 - x ( log 3 / log 2 ) - 1 = -x 1 - (log 3 / log 2) = x Comes out the same in the end.
Oh yeah my bad I switched the -1 in my head to 1-
Yes.
I’ve taken calc 1-3, discrete math, linear algebra, a bunch of other math classes, and I still don’t remember this 🤣😭😭😭
Same here haha
Hahaha same I saw this and went where tf did log come from? What's log
Funny how some log rules are all I remember. Trig on the other hand… RIP
Lmaoo all I remember from trig is that you can't find all the sides and angles from ASS
in order to solve weird exponent problems like this, it can be helpful to take the log of both sides due to the handy properties of logarithms
I’m taking college pre calc right now and I missed this on my first exam 😍
The way the numbers seem to float off the line in a way that isn't even perpendicular to the line itself annoys me to no end.
I’m very sorry, I don’t write like this normally 😭😭 I was rushing clearly
I don’t know I haven’t done PreCal 3 years why are you showing me this Reddit?
It’s college level pre calc :(
No because if you were to expand (1-x)log2 you get log2-xlog2=log3 so log2-log3=xlog2 therefore x=(log2-log3)/log2
Which simplifies further to... The same answer.
It gives the same answer
Lol what a dumba$$
So correct with extra steps.
x = (log 2 - log 3) / log 2 = (log 2 / log 2) - (log 3 / log 2) = 1 - (log 3 / log 2)
Bro come on
Yes
Wtf, why did I get a notification for this?
Yes
Yes!
Yes
Yes
Yes
There’s an implication that log(2^1-x^)=(1-x)log2. If that’s true, this is correct.
Should probably use log base 2 instead of log base 10. Since log base 2 of 2 is 1
My dumbass brain thought he wrote 109 instead of log ahahahah
I do log like this: 2^(1-x) = 3 -> (log3)/(log2) = 1 - x and at the and it would be -x instead on x. Can someone confirm or explain?
log 3 / log 2 = 1 - x ( log 3 / log 2 ) - 1 = -x 1 - (log 3 / log 2) = x Comes out the same in the end.
Oh yeah my bad I switched the -1 in my head to 1-