Question

Can anyone show me the way to solve this one without graphing calculator. 1/4 + sin(pi*x) = 4^(-x). Solve for x. I need help please

Answer 1

This is a tricky question. What are your ideas?

Answer 2

you can start by narrowing down the domain

Answer 3

maybe redefine sin in terms of a polynomial?

Answer 4

or isnt there an e version of sin?

Answer 5

Ok sure. That might work and if it's the right approach it's beyond what I know how to do off the top of my head.

One thing to do, even though you need to find the solution without a graphing calculator is to sketch both functions to see where they might meet. To get an idea of if there are any solutions or just one or many. Just get an idea as to what is going on.

Answer 6

(check your definitions and identities from the course or the chapter ... there are many ways to rewrite functions)

Answer 7

They don't have specific answer and all the equation may beyond my level.

Answer 8

\[4^x(\frac{1}{4} + \sin(pi*x) = 4^{-x})\]
\[4^{x-1} + 4^x\sin(pi*x) = 1\]
is some ways to move things about

Answer 9

here's a picture of the two function on top of each other. so clearly they meet at multiple points.

Answer 10

\[\frac{ 1 }{ 4 } + \sin (pix) = 4^{-x}\]

Answer 11

This is what the question about

Answer 12

\[sin(ab)=sin(a)cos(b)+sin(b)cos(a)\]
\[sin(pi~ x)=sin(pi)cos(x)+sin(x)cos(pi)\]
\[sin(pi~ x)=-sin(x)\]

Answer 13

4^-x is never zero ...

Answer 14

if I have graphing calculator I can see the point where they are intersect. The main idea is I need help to solve for x without graphing calculator mrleiss

Answer 15

I know 4^-x will never be zero

Answer 16

this is a process, you determine what you can do with the knowledge you have at your disposal.

Answer 17

Ok. The other idea is to plug in some values that are special to both exponential functions and trig (or sine in particular) and see if you get any that are the same or close. What values should you try?

Answer 18

This is a free respond question and they don't give you any number to plug in

Answer 19

that trig identity is off, sin (a+b) not ab :/

Answer 20

I know, but what values should you always try - they are the easy ones!

Answer 21

you can't tried any value if you don't know what you are doing

Answer 22

it is definitely not sin(a+b) I swear

Answer 23

You can, and should, think about trying 0, 1, -1 for just about any function or equation; especially if you don't know where to start.

Answer 24

are you in calculus?

Answer 25

It will help you, along with a sketch, to get a feel for what the question is about. You should be able to sketch y = 4^(-x) and y = sin(pi*x) without too much trouble. I put it in the graphing calculator because it's easier online than to sketch in paint.

Answer 26

\[
-1\le\sin(x)\le1\\
-1\le\sin(\pi x)\le1\\
{1\over4}-1\le{1\over4}+\sin(\pi x)\le{1\over4}+1\\
{-3\over4}\le{1\over4}+\sin(\pi x)\le{5\over4}\\
{-3\over4}\le4^{-x}\le{5\over4}\\
\implies 0\le4^{-x}\le{5\over4}\\
-x\ln(4)\le\ln(5)-\ln(4)\\
(1-x)\le\ln(5)
\]

Answer 27

You will still need to use algebra to get at a lot of values of x that are solutions, but it turns out you can get one just by subbing in.

Answer 28

im not sure if there is a pretty way to find "all" values that match

Answer 29

@amistre64 there will be infinite values that "tend" to ge periodic but never are. correct?

Answer 30

thats pretty much that way i see it yes

Answer 31

smallest value for "x"
is still possible to find though

Answer 32

hint-> x must be less than \(\pi/2\)

Answer 33

Well, the answers will approach periodic so you phrase that part of the answer as "for large values, as x -> a + k*pi, sin(x*pi) + 1/4 -> 4^(-x). This should work for all but the first couple of solutions.

Answer 34

@electrokid do you mean x has to be less than 1/2, since we are looking at sin (x*pi)?

Answer 35

yes, x = 0178 something