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A problem Solve for x, 7.3^(x-1)+5-3^(x+1)=0

Mathematics
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\[\huge 7.3^{x-1}+5-3^{x+1}=0\]
is that 7*3^(x-1) or (7.3)^(x-1)
decimal

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Other answers:

hmmm, doesnt look easy
do tell me..does this involve canceling 3^x-1 the solution i thought had that..
\[3^x = \frac{15}{2}\] \[x = \log_3 \frac{15}2\]
@Ishaan94 that's true if the decimal was "dot" *
Oh, my bad... so it's \[(7.3)^{x-1} + 5-3^{x+1}=0\]Hmm Now I feel stupid :/
ok I have an idea
|dw:1333364793236:dw|
|dw:1333364860619:dw|
campbell decimal point, 7.3 = 73/10
:/
its an annoying problem. whats the bloody solution?
ln(a+b) is not the same as lna +lnb
You can't take log like that
I think 7.3 has something to do with 5, we should try to use 5
did sam leave,
wolf's answer http://www.wolframalpha.com/input/?i=7.3%5E%28x-1%29%2B5-3%5E%28x%2B1%29%3D0
why did he say this is easy problem
hhe says its easy and im still stuck on first step, can i switch signs?
Who is hhe?
i meant he, meaning sam
i think this could somehow become a quadratic, with the right substitution
|dw:1333366072740:dw|
i don't think general method we use would work
how do you solve this 2^x - x = 2
I think we could only use newtons method......
first we approximate an answer....
i want an EXACT solution
then use the newton's method to get a better aproximation
we continue this until we're satisfied
but we can't still get the exact though :(
summon sam
currently the best way to solve is graphical method ... lol
yeah
the number 5 is irritating
i sent sam a message, he better respond .
what is classic about this problem anyway?
i assume that is \(7.3^{x-1}\) and not \(7\cdot 3^{x-1}\) so i have no ideas yet
must be something special about 7.3 that i do not see
can you guys tell me later, i gotta sleep
ok try this , let u = 3^(x+1)
i assume there is an exact solution, otherwise youve wasted my time
I can see that you guys think so hard, lol
please give us hint
Sad thing for you @perl There's no exact solution but approximation \[7.3^{x-1}=3^{x+1}-5\]
i knew that newtoms method of approximation yesterday..
I TOLD YOU SO
from yesterday*
it would be more interesting if there was exact solution , :(
i was confused since you said it was easy problem , rats
newton's method :D
newton's method :D
ok then how is it easy to approximate?
Its easy you don't even need to think, lol
what?
it didnt look easy on wolfram
sam , so are you going to post the solution or not. cuz im getting really tired
you can use a calculator for this. is that your 'easy' solution ?
you have to graph it for 7.3^(x-1) and 3^(x+1)-5 pick some points
1 Attachment
what program do you use to graph that?
its a tedious job though, but if you have a graphing calc, it will be fine
then why do you say its easy?
http://www.wolframalpha.com/input/?i=7.3^%28x-1%29%2B5-3^%28x%2B1%29%3D0
what do you mean , pick some points? estimate
yes , estimate
Newtons Method i think..
i dont think so, its just approximation problem
then what else?
anyways, the question seemed pretty ambiguous. i didnt know what they meant by 'easy'
lol, thats all? guessing?
you can use this program to graph https://www.desmos.com/calculator/c
then you do screen capture?
sigh..you can state your question as finding the approximation than 'solve for x'
@Mimi_x3 yes its a bit of disappointing, but I'll come up with a new question later :)
@perl no i dont screen cap on that software
lol, i wanted to see a solution..but there is no solution *sigh*
the solution is approximation ,lol
like the working out..
Nevermind I'll come up a new one :)
Alright, please state your question correctly next time.
so you copy as a file
I think there is a confusion, he copied the problem originally with multiplication 7*3^(x-1)
c'mon i got the approximate thing using 'jugaad' :/

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