Question

2^x-1=3^x+1
solve equation for x , give answer in 2 decimal places.

Answer 1

Whats your guess with first step?

Answer 2

2^x/2=3^x (3)?

Answer 3

This is a fairly challenging problem. To get some orientation regarding what we have to do here, consider graphing f(x)=2^x-1 and g(x)=3^x+1 on the same set of coordinate axes and determining whether or not the two graphs intersect. If they do intersect, you can read an approximate solution right from the graph.

Answer 4

i think no , because it is a log question

Answer 5

This problem is complicated by there being two different bases (2 and 3). What are you going to do about that?

@kY_Tan: You "think no" about what?

Answer 6

about what you saying with the graphing

Answer 7

@kY_Tan; you can not get around this problem with out graphing and since the question says "give answer in 2 decimal places" it has to be an approximation. You have 2 different bases "2" and "3" and you can not get out of it unless you are allowed to use a calculator.

Answer 8

yes, i think this question is allowed to use calculator

Answer 9

the final answer is -4.42

Answer 10

Well if you are allowed to use a calculator in the exam or in such case, then you can solve it.

Answer 11

Why would you consider the graphing suggestion to be inappropriate?

If you don't care to attempt the approximation of a solution by graphing f(x) and g(x) manually, then other options available to you include:

1) Newton's Method
2) Use a graphing calculator's Trace feature to determine an approx. root (x-coordinate of the intersection)
3) Use the "Solver" feature of calculators such as the TI-83 or TI-84

and so on.

Your choice.

Questions?

Answer 12

In regard to your statement, "the final answer is -4.42:" Have you checked this result by substituting it back into the original equation? It's up to you to check every "solution."

Answer 13

@mathmale ; I have seen his previous question and I believe, the examiner wants to test the knowledge of a student about "moveable exponents" and "decimal places", considering he is allowed to use a calculator as he said.

Answer 14

@kY_Tan: what you think would be the the first step?

Answer 15

2^x/2=3^x (3)?

Answer 16

Glad you're coming up with ideas. Mind explaining how this result ( 2^x/2=3^x (3)? ) is going to help you?

Answer 17

because the equation can be simpler by taking off the power?

Answer 18

Simplifying matters is always a good approach, whenever that's possible. Here, I'm afraid you'll need to be a bit more specific in regard to what you'd do to simplify the problem at hand.

Why would you consider the graphing suggestion to be inappropriate?

If you don't care to attempt the approximation of a solution by graphing f(x) and g(x) manually, then other options available to you include:

1) Newton's Method
2) Use a graphing calculator's Trace feature to determine an approx. root (x-coordinate of the intersection)
3) Use the "Solver" feature of calculators such as the TI-83 or TI-84

and so on.

Your choice.

You could also consider whether or not the use of logarithms would help here. I'm not saying that logs would or would not help, but rather posing a suggestion.

Answer 19

@kY_Tan : the key is to think in advance, if simplification will help or not. In this it does not because you if multiply (3*2) you get "6" which can not be written using base "2" or "3" with discrete exponent.

Answer 20

In less than three minutes I was able to graph f(x) and g(x) (as defined above) on my TI-83 Plus calculator, on the same set of axes. This graphing gave me some critically important information. I'd strongly suggest that you, @kY_Tan, graph f(x)=2^x - 1 and g(x) = 3^x + 1 on the same set of coordinate axes, and then describe in words what you observe there.

Answer 21

\[\frac{ 2^x }{ 2 }=3^x(3)\] and if further simplify you get either \[2^x=(6)3^x\]or\[\frac{ 2^x }{ 6 }\]but you can not put "6" in base "3" or "2" with discrete exponent, therefore you should not simply and continue with taking log on both sides.

Answer 22

I would recommend you to follow @mathmale suggestion and draw a graph else the other way would be to take log on both sides and simply.

Answer 23

@imer: Attempting to verify \[\frac{ 2^x }{ 2 }=3^x(3),\]I'm re-writing it as

\[\frac{ 2^x }{ 2^1 }=3^x(3^1),\], or \[2^{x-1}=3^{x+1}\]
Does this equation have a solution? If so, is this solution meaningful in the context of the problem posted by @kY_Tan ?

Answer 24

If you simplify, you will have to come back to the original equation because it makes its make more complicated, and when I said "6" can not be written in base "2" or "3" with discrete exponent, I meant with out splitting "6"

Answer 25

\[\log_{10} 2^{(x-1)}=\log_{10} 3^{(x+1)}\]I am using log with base "10" because all most all calculators have log with base "10". What you think would be the next step? @kY_Tan

Answer 26

@mathmale back in days, I was given such problems and was asked to use log with base "10" by my teacher.

Answer 27

Sounds like an excellent approach! Go ahead and try it. However, don't regard it as "set in stone" that this equation has a solution. Can you either show that it does have one, or that it does not?

Answer 28

x log 2/log 10 = x log (3x10) ? or x-1 log 2 = x+1 log 3?

Answer 29

\[(x-1)\log_{10} (2)=(x+1)\log_{10} (3)\]
Followed by;\[\frac{ x-1 }{ x+1 }=\frac{ \log_{10} (3) }{ \log_{10}(2) }\]

Answer 30

Now find the answer using calculator for right hand side in 2 decimal places and then simplify.

Answer 31

@kY_Tan : what do you get for R.H.S in 2 decimal places?

Answer 32

0.998511482 then should be 1?

Answer 33

no, you get "1.58496....."

Answer 34

you are getting "0.9985..." because you did not put "close bracket" at the end.
It should be "log(3)/log(2)" and not "log(3/log(2"

Answer 35

oh , now i realised

Answer 36

so what is "1.58496..." in 2 decimal places?

Answer 37

1.58?

Answer 38

Correct, now we have;\[\frac{ (x-1) }{ (x+1) }=1.58\]now simplify the equation. What do you get for "x"?

Answer 39

what i get is 1 but i know it is wrong

Answer 40

\[(x-1)=1.58(x+1) \rightarrow x-1=1.58x+1.58 \rightarrow -1-1.58=1.58x-x\]

Answer 41

\[-2.58=0.58x \rightarrow x=\frac{ -2.58 }{ 0.58 }\]

Answer 42

x-1/x+1=1.58
x-1= 1.58x-1.58
x=1.58x-0.58
-0.58x= -0.58
x= 1
this is what i did just now, i made a mistake in multiplying x+1 and 1.58

Answer 43

now i got this but the answer is-4.4482 , if correct to 2 decimal places , will be -4.45 but my textbook answer is -4.42 , is it ok?

Answer 44

We should have taken more than 2 decimal place places before because we want the final answer in 2 decimal places., Therefore we should have taken "1.584" instead of "1.58"

Answer 45

if you take "1.584" instead of "1.58" you will get the final answer as "-4.42" in 2 decimal places.

Answer 46

\[-1-1.584=1.584x-x \rightarrow -2.584=0.584x \rightarrow x=\frac{ -2.584 }{ 0.584 }\]

Answer 47

oh i got the answer now , -4.424 then will be -4.42 if corrected to 2 decimal places

Answer 48

Perfect!

Answer 49

k thanks to you again...

Answer 50

I would recommend you to look for which method is required and confirm with your teacher or friends, either graphing as @mathmale suggested or taking log with the help of calculator.

Answer 51

yes, i think i should do with taking log because i am learning this , while i not seen and learnt the graphing method before

Answer 52

but still thanks to mathmale also

Answer 53

Between by now you should be an expert in solving problems related to log as we have several log problems together :)

Answer 54

i hope so but still have to do more practices