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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

hmmm, doesnt look easy

do tell me..does this involve canceling 3^x-1 the solution i thought had that..

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

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

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

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'

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' :/