.Sam.
A problem
Solve for x,
7.3^(x1)+53^(x+1)=0



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.Sam.
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\[\huge 7.3^{x1}+53^{x+1}=0\]

perl
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is that 7*3^(x1) or (7.3)^(x1)

.Sam.
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decimal

perl
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hmmm, doesnt look easy

lgbasallote
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do tell me..does this involve canceling 3^x1 the solution i thought had that..

Ishaan94
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\[3^x = \frac{15}{2}\]
\[x = \log_3 \frac{15}2\]

.Sam.
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@Ishaan94 that's true if the decimal was "dot" *

Ishaan94
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Oh, my bad... so it's
\[(7.3)^{x1} + 53^{x+1}=0\]Hmm Now I feel stupid :/

perl
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ok I have an idea

perl
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dw:1333364793236:dw

perl
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dw:1333364860619:dw

Ishaan94
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campbell decimal point, 7.3 = 73/10

Ishaan94
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:/

perl
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its an annoying problem. whats the bloody solution?

anonymoustwo44
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ln(a+b) is not the same as lna +lnb

Ishaan94
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You can't take log like that

Ishaan94
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I think 7.3 has something to do with 5, we should try to use 5

perl
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did sam leave,


perl
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why did he say this is easy problem

karatechopper
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hhe says its easy and im still stuck on first step, can i switch signs?

Ishaan94
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Who is hhe?

karatechopper
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i meant he, meaning sam

perl
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i think this could somehow become a quadratic, with the right substitution

perl
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dw:1333366072740:dw

experimentX
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i don't think general method we use would work

experimentX
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how do you solve this 2^x  x = 2

anonymoustwo44
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I think we could only use newtons method......

anonymoustwo44
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first we approximate an answer....

perl
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i want an EXACT solution

anonymoustwo44
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then use the newton's method to get a better aproximation

anonymoustwo44
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we continue this until we're satisfied

anonymoustwo44
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but we can't still get the exact though :(

perl
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summon sam

experimentX
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currently the best way to solve is graphical method ... lol

anonymoustwo44
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yeah

anonymoustwo44
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the number 5 is irritating

perl
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i sent sam a message, he better respond .

perl
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what is classic about this problem anyway?

Ishaan94
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@satellite73

anonymous
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i assume that is \(7.3^{x1}\) and not \(7\cdot 3^{x1}\) so i have no ideas yet

anonymous
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must be something special about 7.3 that i do not see

perl
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can you guys tell me later, i gotta sleep

perl
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ok try this , let u = 3^(x+1)

perl
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i assume there is an exact solution, otherwise youve wasted my time

.Sam.
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I can see that you guys think so hard, lol

perl
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please give us hint

.Sam.
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Sad thing for you @perl
There's no exact solution but approximation
\[7.3^{x1}=3^{x+1}5\]

Mimi_x3
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i knew that newtoms method of approximation yesterday..

anonymoustwo44
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I TOLD YOU SO

Mimi_x3
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from yesterday*

perl
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it would be more interesting if there was exact solution , :(

perl
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i was confused since you said it was easy problem , rats

anonymoustwo44
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newton's method :D

anonymoustwo44
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newton's method :D

perl
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ok then how is it easy to approximate?

.Sam.
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Its easy you don't even need to think, lol

perl
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what?

perl
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it didnt look easy on wolfram

perl
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sam , so are you going to post the solution or not. cuz im getting really tired

perl
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you can use a calculator for this. is that your 'easy' solution ?

.Sam.
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you have to graph it for 7.3^(x1) and 3^(x+1)5 pick some points

perl
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what program do you use to graph that?

.Sam.
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its a tedious job though, but if you have a graphing calc, it will be fine

perl
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then why do you say its easy?


perl
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what do you mean , pick some points? estimate

.Sam.
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yes , estimate

Mimi_x3
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Newtons Method i think..

perl
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i dont think so, its just approximation problem

Mimi_x3
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then what else?

perl
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anyways, the question seemed pretty ambiguous. i didnt know what they meant by 'easy'

Mimi_x3
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lol, thats all? guessing?


perl
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then you do screen capture?

Mimi_x3
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sigh..you can state your question as finding the approximation than 'solve for x'

.Sam.
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@Mimi_x3 yes its a bit of disappointing, but I'll come up with a new question later :)

.Sam.
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@perl no i dont screen cap on that software

Mimi_x3
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lol, i wanted to see a solution..but there is no solution *sigh*

.Sam.
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the solution is approximation ,lol

Mimi_x3
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like the working out..

.Sam.
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Nevermind I'll come up a new one :)

Mimi_x3
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Alright, please state your question correctly next time.

perl
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so you copy as a file

perl
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I think there is a confusion, he copied the problem originally with multiplication
7*3^(x1)

apoorvk
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c'mon i got the approximate thing using 'jugaad' :/