Question

How do I solve this algerbraically? (made up question)

Answer 1

\(\large\color{black}{ \displaystyle 3(2)^x=15x-1 }\)

Answer 2

(If I will need to know calculus, then please, because I know (some) calculus)

Answer 3

try taking logarithms

Answer 4

\(\large\color{black}{ \displaystyle x\ln(2)=\ln\left(x-\frac{1}{3}\right) }\)

Answer 5

I divided by 3 first, if that is any bettter.

Answer 6

made log base 2?

Answer 7

maybe**

Answer 8

\[3.2^x=15x-1\]
is not an expression solvable with algebra, what I would suggest you do is create two functions:
\[u(x)=3.2^x\]
\[f(x)=15x-1\]
Graph them using a software or a calculator and then find the intersection.

Answer 9

Yes, that is what I would do normally, but is there not an algebra (maybe involving calculus) solution?

Answer 10

Not that I know about at least.
I have already tried solving these on my free time, and never got to the answer, maybe I a not good enough with math.

Answer 11

yes - i couldn't get anywhere using logs

Answer 12

I have seen people say that these are solvabe ... maybe I can try power series or something

Answer 13

I tried that, it is hard to accomodate the function in order to apply it.

Answer 14

\(\large\color{black}{ \displaystyle 1=\ln(2)x^{-1}\ln\left(x-\frac{1}{3}\right) }\)

\(\large\color{black}{ \displaystyle \ln(2)x^{-1} \ln\left(x-\frac{1}{3}\right)=\ln(2)\sum_{i=0}^{\infty}\frac{x^{-1}(-1)^i(-4/3{~~}+x)^i}{i} }\)

Answer 15

Yeah, just a time waster ...

Answer 16

seies for ln(2) is better :O

Answer 17

I'd immediately try Newton's Method. Would y ou consider that to be an "algebraic" approach?

Answer 18

Stepsize?

Answer 19

What I haven't tried yet is to use the notable equality \[e^{\ln(f(x))}=f(x)\]

Answer 20

The only think in the Newton related to math I have hear is, that stepsize formula for approximating differential equation-solutions.

Answer 21

heard**

Answer 22

right! Newtons method could be applicable here
That is algebraic

Answer 23

\(\large\color{black}{ \displaystyle x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} }\)

are you referring to this?

Answer 24

Indeed I am. Very good!

Answer 25

So the question here is: How do we make a function f(x) out of 3(2)^x=15x-1?

Answer 26

yes you start with an approximate root
and further approximations are made as follows

x(n+1) = x(n) - f( x(n) / f' x(n)

Answer 27

This'll be interesting to know.

Answer 28

\(\large\color{black}{ \displaystyle 3(2)^x=15x-1 }\)

\(\large\color{black}{ \displaystyle (2)^x=5x-1/3 }\)

\(\large\color{black}{ \displaystyle x=\frac{(\ln(5x-1/3)}{\ln(2)} }\)

Answer 29

\(\large\color{black}{ \displaystyle 1=\frac{(\ln(5x-1/3)}{x\ln(2)} }\)

Answer 30

\(\large\color{black}{ \displaystyle f(x)=\frac{(\ln(5x-1/3)}{x\ln(2)} -1}\)


ike this?

Answer 31

Next suggestion: It'd be nice to find a ballpark approx. for x. You could do that by graphing 3(x)^x and 15-1 x on the same set of axes and determining the x value at which the two graphs intersect.
Hate to say "no" to your suggestion, but no. If 3(2)^x = 15 - 1, rewrite this as f(x) = 3(2)^x -15x + 1.

Answer 32

Oh, I am so stupid!
THanks :)

Ok, I will graph the 2^x and 15x-1....

Answer 33

Good idea. That way you can obtain a starting value for x.

Answer 34

but that wouldn't be algebraic would it?

Answer 35

https://www.desmos.com/calculator/g6uh2od1ya

Answer 36

Owlcoffee I think that is just to check if it will make sense

Answer 37

reasonability check

Answer 38

Graph 3(2)^x, not 2^x.
No, it wouldn't be "algbraic," BUT the purpose of doing this is to obtain a starting value for your use of Newton's Method.

Answer 39

Newton's Method is both algebraic and an application of calculus.

Answer 40

well, my problem was
\(\large\color{black}{ \displaystyle 3(2)^x=15x-1 }\)

Answer 41

Yes, that's the problem I was addressing also.

Answer 42

So why 2^x and not 3(2)^x ?

Answer 43

here is my reasoning:
we can rewrite your equation as follows:
\[\huge {2^x} = 5x - \frac{1}{3}\]
now, we can write the exponential function, using the Taylor's expansion around \(x=0\), so we get:
\[\huge {2^x} \cong 1 + x\ln 2 + {x^2}\frac{{{{\left( {\ln 2} \right)}^2}}}{2}\]

Answer 44

I am saying you need to graph / use 3(2)^x, not (2)^x alone. Refer to your original problem statement.

Answer 45

I did graph 3(2)^x, didn't I ?

Answer 46

Maybe a change of variable can do some good?... Well no... isolating the "x" would yield some trouble...

Answer 47

Anyway, now I will repost the link ...
https://www.desmos.com/calculator/g6uh2od1ya

Answer 48

there's more than one way to skin a cat. Michelle_Laino's suggestion is fine.

Answer 49

thanks!! :) @mathmale

Answer 50

Yeah and then solve a quandratic (with wierd coefficients, but doesn't matter...)
Thanks Michele Laino !

Answer 51

Not algebraic, but satisfies me totally

Answer 52

Again, the equation we want to solve for x is 3(2)^x=15x-1. Alternatively, solve the following for x: 2^x = 3x - 1/3. Your choice. This IS algebraic.

Answer 53

:)
maybe it is suffice to stop at the first order term in Taylor's expansion @idku

Answer 54

Whatever, a quadratic poly is also fine. In that case even a cubic would be.

Answer 55

I don't know how much less accurate the solution is going to be if you stop at n=1 degree, but I don't think we really need to do this though...

Answer 56

Graph the following:\[y _{1}=2^x,y _{2}=5x-1/3.\]

Answer 57

2^x=5x-1, you meant mathmale?

Answer 58

oh, :D you got me .. o

Answer 59

yes! you can try to consider an expansion of the third order @idku

Answer 60

Alright

https://www.desmos.com/calculator/xnhohek0om

Answer 61

You, idku, need to decide for yourself the degree of accuracy you want. Either a Taylor Series expansion or the use of Newton"s Method will produce whatever degree of accuracy you want.

Answer 62

Why not say: "I want my root accurate to three decimal places?"

Answer 63

Well, it would be nice to know everything I can, even though I am not even close to be close to be close etc... to a mathematician.

Answer 64

You posed this problem, so you have the prerogative of specifying the level of accuracy to which y ou want to find a solution.

Answer 65

I would think that a cubic polynomial would provide me 4 decimal places of accurace.... just need to graph it and see how smoothly it lies on the solution.... (I suppose)

Answer 66

just enjoy the experience!

Answer 67

yes., that is what I am trying to do :)

Answer 68

Let me settle the argument (if there is an argument here): Solve 3(2)^x=15x-1 to 2 decimal place accuracy.

Answer 69

I just haven't really done the Newton's method which you sugessted... can you direct me and I will do my best?

Answer 70

(this isn't my h/w anyway, but no direct answers policy must be respected)

Answer 71

First, please graph the 2 functions and estimate the x-coordinate of the solution.

Answer 72

\(\large\color{black}{ \displaystyle 2^x=5x-\frac{1}{3} }\)


\(\large\color{black}{ \displaystyle y_1=5x-\frac{1}{3} }\) \(\large\color{black}{ \displaystyle y_2=2^x }\)

\(\large\color{black}{ \displaystyle x=0.316 }\) (via desmos)

Answer 73

x=0.32
(to two decimal places)

Answer 74

I'll take your word for it. Let your starting value be .32 or just .3. Close enough!
Next, write out f(x) as explained before.

Answer 75

\(\large\color{black}{ \displaystyle f(x)=2^x+5x-\frac{1}{3} }\)

(would impact only the y-value where this solution lays, but the solution for x is same when dividing both functions by 3)

Answer 76

So I first lug in x=0.3 into the f(x) ?

Answer 77

plug

Answer 78

yes. In other words, calculate f(.3).
Next, take the derivative of f(x). What is it?
Next, evaluate the derivative at x=.3.

Answer 79

\(\large\color{black}{ \displaystyle f(x)=2^x+5x-\frac{1}{3} }\)

\(\large\color{black}{ \displaystyle f'(x)=2^x\ln(2)+5 }\)


\(\large\color{black}{ \displaystyle f'(0.3)=2^{0.3}\ln(0.3)+5=\\[0.7em] 1.23\cdot (-1.20)+5=3.524}\)

Answer 80

oh, my ln(0.3) i am dump

Answer 81

Now insert your results into the following formula for the approximate root:

\[x _{1}=x _{0}-\frac{ f(x_0) }{ f'(x _{0}) }\]

Answer 82

\(\large\color{black}{ \displaystyle f'(0.3)=2^{0.3}\ln(2)+5=\\[0.7em] 1.23\cdot (0.69)+5=5.8487\approx 5.85}\)

Answer 83

\(\large\color{black}{ \displaystyle x_1=0.3-\frac{2.40}{5.85}}\)

Answer 84

I evaluated f(0.3) as well...

Answer 85

So you'll be calculating \[x _{1}=0.3-\frac{ f(0.3) }{ f'(0.3) }\]

Answer 86

yes.

Answer 87

Please do this work. Your result should be somewhere close to x_1 = 0.3.

Answer 88

-0.11

Answer 89

or, -0.109647...

Answer 90

that doesn't agree with our starting x value, 0.3. Arithmetic mistake likely.

Answer 91

http://www.wolframalpha.com/input/?i=0.3-%282%5E%280.3%29%2B5%280.3%29-1%2F3%29%2F%282%5E%280.3%29ln%282%29%2B5%29

Answer 92

everything seems to be correctly plugged.

Answer 93

i cheated and use an equation calculator which gave a value of x = 0.316

Answer 94

lol, that is what desmos.com gives right off...

Answer 95

yes

Answer 96

I don't want to say that I just want to know how to do it, because it is too banal, but really just want to know how to do it...

Answer 97

unfortunate result ',',',',',