# 5X = (1+X)^15 ; solve for X

#### jove8414

5X = (1+X)^15

What to do with this kinds of problems? How do I solve for X? I encounter this problems on engineering economy(Annuities)... I tried doing logarithms but no success.

#### BvU

Homework Helper
Hi Jove,

What did you do to try logarithms ? Please post so we can see where things go wrong ...
However, perhaps a numerical approach is in order for this kind of exercise (e.g. if x should be a natural number).

#### PeroK

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It would be difficult to get. get a closed form solution. You will need to do a numerical approximation.
You have a polynomial of degree 15.

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#### jove8414

Hi Jove,

What did you do to try logarithms ? Please post so we can see where things go wrong ...
However, perhaps a numerical approach is in order for this kind of exercise (e.g. if x should be a natural number).
I tried changing the equations into logarithm form:
log(5X) = 15log(1+X)

and afterwards i still can't solve for X

You won't get a closed form solution. You will need to do a numerical approximation.
Yeah the book also did the same (numerical approximation). I thought there's a way(s) to do it other than that :/

#### BvU

Homework Helper
log(5X) = 15log(1+X)
one more step gets $\log 5 + \log x = 15 \log (1+x)$ and then it is indeed hopeless. So either a numerical approach or a graphical solution is the only way out.

#### PeroK

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2018 Award
$u = x +1$ is an obvious substitution. There must be an algebraic solution but I don't know any advanced techniques for so high a power.

#### BvU

Homework Helper
Is there a solution at all ? If I look at the numbers I don't think so !
Or at the graph

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#### HallsofIvy

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$u = x +1$ is an obvious substitution. There must be an algebraic solution but I don't know any advanced techniques for so high a power.
No "must" about it. There exist "algebraic solutions" for polynomial equations of degree 4 or less but there is no general "algebraic solution" for such equations of degree 5 or higher.

And the substitution u= x+ 1 just "shifts" the problem. You have $u^{15}= 5(u- 1)$, still a polynomial equation of degree 15.

#### nasu

Is there a solution at all ? If I look at the numbers I don't think so !
Or at the graph
By taking the log you assume that x+1>0 and eliminate solutions less than -1.

The same Wolfram Alpha will give a real solution of about -2.17, for the initial problem.

BvU

#### BvU

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I wonder how this answer fits in the annuity context, though

#### BvU

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So jove, is your exercise really 5X = (1+X)^15 or does it ask what interest rate you need to have five times your initial deposit after 15 years ?

#### PeroK

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2018 Award
No "must" about it. There exist "algebraic solutions" for polynomial equations of degree 4 or less but there is no general "algebraic solution" for such equations of degree 5 or higher.

And the substitution u= x+ 1 just "shifts" the problem. You have $u^{15}= 5(u- 1)$, still a polynomial equation of degree 15.
The solutions to this equation must be algebraic, by definition.

#### jbriggs444

Homework Helper
What definition?
An "algebraic number" is one that is a solution of a non-zero polynomial equation in one variable with integer coefficients. Real numbers that are not "algebraic" are, instead, "transcendental".

However, these "algebraic numbers" may not be expressible in terms of elementary operations -- addition, subtraction, multiplication, division and root extraction, hence the Abel Ruffini theorem.

#### PeroK

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2018 Award
I meant algebraic numbers. And I was wondering about techniques to solve a particular equation. You don't necessarily need a general solution to every polynomial of order 15 in order to solve this one.

#### pbuk

5X = (1+X)^15

What to do with this kinds of problems? How do I solve for X? I encounter this problems on engineering economy(Annuities)... I tried doing logarithms but no success.
I wonder how this answer fits in the annuity context, though
It does not of course; there appears to be a typo in the OP. A problem which does makes sense in the context of annuities (although it is a simple compound interest problem) is $5 = (1+x)^{15}$ which is easily solved for its unique real solution.

#### BvU

Homework Helper
I'd like to hear that from Jove himself (herself?) ...
(as in: post #11)

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