How do you solve
(x^3)+x-16=0 ?
I was solving an exponential equation and it came to this and I am stuck. Calculators can produce the answers but I want to get the working. Its been getting me worried for the past 5 days now.
Original Question
(2^x)+2^(3x)=16

Post the original question

meum:
Post the original question

Done.

It's been a long time I did all this algebra gymnastics.

Why aren't you doing something like
2^x+2^(3x)=2^4?

And simplify as they have similar base? Na 20+ years ago I do these kind things last

X na 1

Is the solution from the calculator 1.2556? Let me know so I can share the step by step working

meum:
Is the solution from the calculator 1.2556? Let me know so I can share the step by step working

If you're confident of your answer you won't be posting this.

Unless you have no idea what you're solving.

Sleekcode:


If you're confident of your answer you won't be posting this.

Unless you have no idea what you're solving.

Why una like fight like dis for dis forum & the country sef?

Oga carry pen & paper. Question dey dia. Solve your own! E no hard.

Allow make d poster answer me.

No be today I begin solve maths problem for this forum.

meum:
Is the solution from the calculator 1.2556?

Yes. If I had used casio natural display maybe I would have analysed the answer in surd form and try to work it back .

airsaylongcome:
It's been a long time I did all this algebra gymnastics.

Why aren't you doing something like
2^x+2^(3x)=2^4?

And simplify as they have similar base? Na 20+ years ago I do these kind things last

I tried this but I couldn't get farther than that so I decided to factorize. When I factorized the indices and substituted an arbitrary letter 'x' for 2x, the cubic equation became a problem too

That's the answer I got.

BenedictAbajue:
Yes. If I had used casio natural display maybe I would have analysed the answer in surd form and try to work it back .

Okay. I will send in the step by step solution soon. Let me make it visible enough.😂

(2^x) + 2^(3x) = 16

ie. 2^x + 2^x*3 = 16 ……eqn 1

Now let 2^x = y ….. eqn 2

Eqn 1 becomes
y + y^3 = 16

rearranging this you will have:

y^3 + y - 16 = 0 …….eqn 3

The above, equation 2 is in the form

y^3 + py+ q = 0 ……eqn 4

Applying Cardano’s formula below
y=(3√−q/2+√q^2/4+p^3/27) + (3√−q/2−√q^2/4+p^3/27).

Comparing eqns 3 & 4 then substituting the values for p & q you’ll have:

y = 2.387 (plug in the values yourself)

Now from eqn 2,
2^x = y

This implies that 2^x = 2.387

Taking log of both sides, you will have

log 2^x = log 2.387

This implies that:
xlog2 = log2.387

x = log2.387 / log2

Therefore x = 1.2556 👍


Ask questions where you don’t understand.

Note the 3√ in cardona’s formula covers the entire values within each bracket() & it means cube root not 3*√

By log I imply log to base ten(10) of both sides

meum:

Okay. I will send in the step by step solution soon. Let me rewrite and make it visible enough

Alright

platymus:
That's the answer I got.

You can't add up their index like that unless it was multiplication instead of addition that was between them

platymus:
That's the answer I got.

Nah. Maths rule was broken.

meum:


Ask questions where you don’t understand

This is the first time I an hearing of Cardano's formula. So the formula works for all equations in that form. I was hoping to see a direct solving of the cubic equation itself. What if the question was something like this:
(2^x)+3^(3x)=16

BenedictAbajue:

This is the first time I an hearing of Cardano's formula. So the formula works for all equations in that form. I was hoping to see a direct solving of the cubic equation itself. What if the question was something like this:
(2^x)+3^(3x)=16

The Cardano's formula (named after Girolamo Cardano 1501-1576), which is similar to the perfect-square method to quadratic equations, is a standard way to find a real root of a cubic equation.

If the question were written as above then it will be mathematically incorrect😂

meum:


The Cardano's formula (named after Girolamo Cardano 1501-1576), which is similar to the perfect-square method to quadratic equations, is a standard way to find a real root of a cubic equation.

If the question were written as above then it will be mathematically incorrect😂

Nice!

Since this is a programming section, you have to solve the the programming way using something like numerical analysis..

You could start from 0 to maybe like a million and input them as value of x to see if it equat zero then break. Lol computers would do this very fast. You could also try negative numbers too.

Answer can also be decimal you could also check for that too.

meum:


If the question were written as above then it will be mathematically incorrect

'Incorrect' like there is no answer or that the formula wont work on it. Which one?

Deicide:
Since this is a programming section, you have to solve the the programming way using something like numerical analysis..

You could start from 0 to maybe like a million and input them as value of x to see if it equat zero then break. Lol computers would do this very fast. You could also try negative numbers too.

Answer can also be decimal you could also check for that too.

I could do this with python but I need the solvings.

platymus:
That's the answer I got.

This was how we were taught in secondary school, so anybody screaming Cardano or Cardoso or Ethereum formula take note, doesn’t this also arrive at the same answer without stress??

2paid2Bmad:


This was how we were taught in secondary school, so anybody screaming Cardano or Cardoso or Ethereum formula take note, doesn’t this also arrive at the same answer without stress??

Oya put 1 into the equation and check if you are correct na

BenedictAbajue:
I tried this but I couldn't get farther than that so I decided to factorize. When I factorized the indices and substituted an arbitrary letter 'x' for 2x, the cubic equation became a problem too

2^x + 2^3x = 16 =
2^x + 2^x+2x = 16 =
2^x (1+2^2x) = 16 =
1 + 2^(2x) = 16 / 2^x =
1 + 2^(2x) = 2^4 / 2^x =
1 + 2^(2x) = 2^(4-x) =
2^(2x) = 2^(4-x) - 1 =
2x = log2(2^(4-x) - 1) =
2x = log2(2^(4-x)) + log2(1 - 1/2^(4-x)) =
2x = (4-x)log2(2) + log2(2^(4-x) - 1) =
x = (4 + log2(2^(4-x) - 1)) / 3

studyless123:

2^x + 2^3x = 16 = 2^x + 2^x+2x = 16 = 2^x (1+2^2x) = 16 = 1 + 2^(2x) = 16/2^x

From this point, it went all wrong.

(x^3)+x-16=0.
(x^3)+(0x^2)+x-16=0.
(((x+0)x+1)x-16).
Use trial method (random consecutive numbers in place of "x"), when the trinomial equals zero, that is one solution or root of the equation.
You then do polynomial division to get the remainder which is a quadratic equation and solve for the remaining roots.

platymus:
That's the answer I got.

Wrong answer