.Sam.
  • .Sam.
A problem Solve for x, 7.3^(x-1)+5-3^(x+1)=0
Mathematics
katieb
  • katieb
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.Sam.
  • .Sam.
\[\huge 7.3^{x-1}+5-3^{x+1}=0\]
perl
  • perl
is that 7*3^(x-1) or (7.3)^(x-1)
.Sam.
  • .Sam.
decimal

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perl
  • perl
hmmm, doesnt look easy
lgbasallote
  • lgbasallote
do tell me..does this involve canceling 3^x-1 the solution i thought had that..
anonymous
  • anonymous
\[3^x = \frac{15}{2}\] \[x = \log_3 \frac{15}2\]
.Sam.
  • .Sam.
@Ishaan94 that's true if the decimal was "dot" *
anonymous
  • anonymous
Oh, my bad... so it's \[(7.3)^{x-1} + 5-3^{x+1}=0\]Hmm Now I feel stupid :/
perl
  • perl
ok I have an idea
perl
  • perl
|dw:1333364793236:dw|
perl
  • perl
|dw:1333364860619:dw|
anonymous
  • anonymous
campbell decimal point, 7.3 = 73/10
anonymous
  • anonymous
:/
perl
  • perl
its an annoying problem. whats the bloody solution?
anonymous
  • anonymous
ln(a+b) is not the same as lna +lnb
anonymous
  • anonymous
You can't take log like that
anonymous
  • anonymous
I think 7.3 has something to do with 5, we should try to use 5
perl
  • perl
did sam leave,
experimentX
  • experimentX
wolf's answer http://www.wolframalpha.com/input/?i=7.3%5E%28x-1%29%2B5-3%5E%28x%2B1%29%3D0
perl
  • perl
why did he say this is easy problem
karatechopper
  • karatechopper
hhe says its easy and im still stuck on first step, can i switch signs?
anonymous
  • anonymous
Who is hhe?
karatechopper
  • karatechopper
i meant he, meaning sam
perl
  • perl
i think this could somehow become a quadratic, with the right substitution
perl
  • perl
|dw:1333366072740:dw|
experimentX
  • experimentX
i don't think general method we use would work
experimentX
  • experimentX
how do you solve this 2^x - x = 2
anonymous
  • anonymous
I think we could only use newtons method......
anonymous
  • anonymous
first we approximate an answer....
perl
  • perl
i want an EXACT solution
anonymous
  • anonymous
then use the newton's method to get a better aproximation
anonymous
  • anonymous
we continue this until we're satisfied
anonymous
  • anonymous
but we can't still get the exact though :(
perl
  • perl
summon sam
experimentX
  • experimentX
currently the best way to solve is graphical method ... lol
anonymous
  • anonymous
yeah
anonymous
  • anonymous
the number 5 is irritating
perl
  • perl
i sent sam a message, he better respond .
perl
  • perl
what is classic about this problem anyway?
anonymous
  • anonymous
@satellite73
anonymous
  • anonymous
i assume that is \(7.3^{x-1}\) and not \(7\cdot 3^{x-1}\) so i have no ideas yet
anonymous
  • anonymous
must be something special about 7.3 that i do not see
perl
  • perl
can you guys tell me later, i gotta sleep
perl
  • perl
ok try this , let u = 3^(x+1)
perl
  • perl
i assume there is an exact solution, otherwise youve wasted my time
.Sam.
  • .Sam.
I can see that you guys think so hard, lol
perl
  • perl
please give us hint
.Sam.
  • .Sam.
Sad thing for you @perl There's no exact solution but approximation \[7.3^{x-1}=3^{x+1}-5\]
Mimi_x3
  • Mimi_x3
i knew that newtoms method of approximation yesterday..
anonymous
  • anonymous
I TOLD YOU SO
Mimi_x3
  • Mimi_x3
from yesterday*
perl
  • perl
it would be more interesting if there was exact solution , :(
perl
  • perl
i was confused since you said it was easy problem , rats
anonymous
  • anonymous
newton's method :D
anonymous
  • anonymous
newton's method :D
perl
  • perl
ok then how is it easy to approximate?
.Sam.
  • .Sam.
Its easy you don't even need to think, lol
perl
  • perl
what?
perl
  • perl
it didnt look easy on wolfram
perl
  • perl
sam , so are you going to post the solution or not. cuz im getting really tired
perl
  • perl
you can use a calculator for this. is that your 'easy' solution ?
.Sam.
  • .Sam.
you have to graph it for 7.3^(x-1) and 3^(x+1)-5 pick some points
1 Attachment
perl
  • perl
what program do you use to graph that?
.Sam.
  • .Sam.
its a tedious job though, but if you have a graphing calc, it will be fine
perl
  • perl
then why do you say its easy?
perl
  • perl
http://www.wolframalpha.com/input/?i=7.3^%28x-1%29%2B5-3^%28x%2B1%29%3D0
perl
  • perl
what do you mean , pick some points? estimate
.Sam.
  • .Sam.
yes , estimate
Mimi_x3
  • Mimi_x3
Newtons Method i think..
perl
  • perl
i dont think so, its just approximation problem
Mimi_x3
  • Mimi_x3
then what else?
perl
  • perl
anyways, the question seemed pretty ambiguous. i didnt know what they meant by 'easy'
Mimi_x3
  • Mimi_x3
lol, thats all? guessing?
.Sam.
  • .Sam.
you can use this program to graph https://www.desmos.com/calculator/c
perl
  • perl
then you do screen capture?
Mimi_x3
  • Mimi_x3
sigh..you can state your question as finding the approximation than 'solve for x'
.Sam.
  • .Sam.
@Mimi_x3 yes its a bit of disappointing, but I'll come up with a new question later :)
.Sam.
  • .Sam.
@perl no i dont screen cap on that software
Mimi_x3
  • Mimi_x3
lol, i wanted to see a solution..but there is no solution *sigh*
.Sam.
  • .Sam.
the solution is approximation ,lol
Mimi_x3
  • Mimi_x3
like the working out..
.Sam.
  • .Sam.
Nevermind I'll come up a new one :)
Mimi_x3
  • Mimi_x3
Alright, please state your question correctly next time.
perl
  • perl
so you copy as a file
perl
  • perl
I think there is a confusion, he copied the problem originally with multiplication 7*3^(x-1)
apoorvk
  • apoorvk
c'mon i got the approximate thing using 'jugaad' :/

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