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

- .Sam.

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

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- .Sam.

\[\huge 7.3^{x-1}+5-3^{x+1}=0\]

- perl

is that 7*3^(x-1) or (7.3)^(x-1)

- .Sam.

decimal

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## More answers

- perl

hmmm, doesnt look easy

- lgbasallote

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

- anonymous

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

- .Sam.

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

- anonymous

Oh, my bad... so it's
\[(7.3)^{x-1} + 5-3^{x+1}=0\]Hmm Now I feel stupid :/

- perl

ok I have an idea

- perl

|dw:1333364793236:dw|

- perl

|dw:1333364860619:dw|

- anonymous

campbell decimal point, 7.3 = 73/10

- anonymous

:/

- perl

its an annoying problem. whats the bloody solution?

- anonymous

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

- anonymous

You can't take log like that

- anonymous

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

- perl

did sam leave,

- experimentX

wolf's answer
http://www.wolframalpha.com/input/?i=7.3%5E%28x-1%29%2B5-3%5E%28x%2B1%29%3D0

- perl

why did he say this is easy problem

- karatechopper

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

- anonymous

Who is hhe?

- karatechopper

i meant he, meaning sam

- perl

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

- perl

|dw:1333366072740:dw|

- experimentX

i don't think general method we use would work

- experimentX

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

- anonymous

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

- anonymous

first we approximate an answer....

- perl

i want an EXACT solution

- anonymous

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

- anonymous

we continue this until we're satisfied

- anonymous

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

- perl

summon sam

- experimentX

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

- anonymous

yeah

- anonymous

the number 5 is irritating

- perl

i sent sam a message, he better respond .

- perl

what is classic about this problem anyway?

- anonymous

@satellite73

- anonymous

i assume that is \(7.3^{x-1}\) and not \(7\cdot 3^{x-1}\) so i have no ideas yet

- anonymous

must be something special about 7.3 that i do not see

- perl

can you guys tell me later, i gotta sleep

- perl

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

- perl

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

- .Sam.

I can see that you guys think so hard, lol

- perl

please give us hint

- .Sam.

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

- Mimi_x3

i knew that newtoms method of approximation yesterday..

- anonymous

I TOLD YOU SO

- Mimi_x3

from yesterday*

- perl

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

- perl

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

- anonymous

newton's method :D

- anonymous

newton's method :D

- perl

ok then how is it easy to approximate?

- .Sam.

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

- perl

what?

- perl

it didnt look easy on wolfram

- perl

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

- perl

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

- .Sam.

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

##### 1 Attachment

- perl

what program do you use to graph that?

- .Sam.

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

- perl

then why do you say its easy?

- perl

http://www.wolframalpha.com/input/?i=7.3^%28x-1%29%2B5-3^%28x%2B1%29%3D0

- perl

what do you mean , pick some points? estimate

- .Sam.

yes , estimate

- Mimi_x3

Newtons Method i think..

- perl

i dont think so, its just approximation problem

- Mimi_x3

then what else?

- perl

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

- Mimi_x3

lol, thats all? guessing?

- .Sam.

you can use this program to graph
https://www.desmos.com/calculator/c

- perl

then you do screen capture?

- Mimi_x3

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

- .Sam.

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

- .Sam.

@perl no i dont screen cap on that software

- Mimi_x3

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

- .Sam.

the solution is approximation ,lol

- Mimi_x3

like the working out..

- .Sam.

Nevermind I'll come up a new one :)

- Mimi_x3

Alright, please state your question correctly next time.

- perl

so you copy as a file

- perl

I think there is a confusion, he copied the problem originally with multiplication
7*3^(x-1)

- apoorvk

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

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