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

This Question is Closed

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4\[\huge 7.3^{x1}+53^{x+1}=0\]

perl
 4 years ago
Best ResponseYou've already chosen the best response.0is that 7*3^(x1) or (7.3)^(x1)

lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0do tell me..does this involve canceling 3^x1 the solution i thought had that..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[3^x = \frac{15}{2}\] \[x = \log_3 \frac{15}2\]

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4@Ishaan94 that's true if the decimal was "dot" *

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh, my bad... so it's \[(7.3)^{x1} + 53^{x+1}=0\]Hmm Now I feel stupid :/

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0campbell decimal point, 7.3 = 73/10

perl
 4 years ago
Best ResponseYou've already chosen the best response.0its an annoying problem. whats the bloody solution?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ln(a+b) is not the same as lna +lnb

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You can't take log like that

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think 7.3 has something to do with 5, we should try to use 5

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0wolf's answer http://www.wolframalpha.com/input/?i=7.3%5E%28x1%29%2B53%5E%28x%2B1%29%3D0

perl
 4 years ago
Best ResponseYou've already chosen the best response.0why did he say this is easy problem

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0hhe says its easy and im still stuck on first step, can i switch signs?

karatechopper
 4 years ago
Best ResponseYou've already chosen the best response.0i meant he, meaning sam

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i think this could somehow become a quadratic, with the right substitution

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0i don't think general method we use would work

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0how do you solve this 2^x  x = 2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think we could only use newtons method......

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0first we approximate an answer....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0then use the newton's method to get a better aproximation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we continue this until we're satisfied

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but we can't still get the exact though :(

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.0currently the best way to solve is graphical method ... lol

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the number 5 is irritating

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i sent sam a message, he better respond .

perl
 4 years ago
Best ResponseYou've already chosen the best response.0what is classic about this problem anyway?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i assume that is \(7.3^{x1}\) and not \(7\cdot 3^{x1}\) so i have no ideas yet

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0must be something special about 7.3 that i do not see

perl
 4 years ago
Best ResponseYou've already chosen the best response.0can you guys tell me later, i gotta sleep

perl
 4 years ago
Best ResponseYou've already chosen the best response.0ok try this , let u = 3^(x+1)

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i assume there is an exact solution, otherwise youve wasted my time

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4I can see that you guys think so hard, lol

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4Sad thing for you @perl There's no exact solution but approximation \[7.3^{x1}=3^{x+1}5\]

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0i knew that newtoms method of approximation yesterday..

perl
 4 years ago
Best ResponseYou've already chosen the best response.0it would be more interesting if there was exact solution , :(

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i was confused since you said it was easy problem , rats

perl
 4 years ago
Best ResponseYou've already chosen the best response.0ok then how is it easy to approximate?

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4Its easy you don't even need to think, lol

perl
 4 years ago
Best ResponseYou've already chosen the best response.0it didnt look easy on wolfram

perl
 4 years ago
Best ResponseYou've already chosen the best response.0sam , so are you going to post the solution or not. cuz im getting really tired

perl
 4 years ago
Best ResponseYou've already chosen the best response.0you can use a calculator for this. is that your 'easy' solution ?

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4you have to graph it for 7.3^(x1) and 3^(x+1)5 pick some points

perl
 4 years ago
Best ResponseYou've already chosen the best response.0what program do you use to graph that?

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4its a tedious job though, but if you have a graphing calc, it will be fine

perl
 4 years ago
Best ResponseYou've already chosen the best response.0then why do you say its easy?

perl
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=7.3^%28x1%29%2B53^%28x%2B1%29%3D0

perl
 4 years ago
Best ResponseYou've already chosen the best response.0what do you mean , pick some points? estimate

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0Newtons Method i think..

perl
 4 years ago
Best ResponseYou've already chosen the best response.0i dont think so, its just approximation problem

perl
 4 years ago
Best ResponseYou've already chosen the best response.0anyways, the question seemed pretty ambiguous. i didnt know what they meant by 'easy'

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0lol, thats all? guessing?

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4you can use this program to graph https://www.desmos.com/calculator/c

perl
 4 years ago
Best ResponseYou've already chosen the best response.0then you do screen capture?

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0sigh..you can state your question as finding the approximation than 'solve for x'

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4@Mimi_x3 yes its a bit of disappointing, but I'll come up with a new question later :)

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4@perl no i dont screen cap on that software

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0lol, i wanted to see a solution..but there is no solution *sigh*

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4the solution is approximation ,lol

.Sam.
 4 years ago
Best ResponseYou've already chosen the best response.4Nevermind I'll come up a new one :)

Mimi_x3
 4 years ago
Best ResponseYou've already chosen the best response.0Alright, please state your question correctly next time.

perl
 4 years ago
Best ResponseYou've already chosen the best response.0I think there is a confusion, he copied the problem originally with multiplication 7*3^(x1)

apoorvk
 4 years ago
Best ResponseYou've already chosen the best response.0c'mon i got the approximate thing using 'jugaad' :/
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