0

For Example: $5+5x=4\cdot 1.03^x$

My Question is: Is there a mathematical way that lets me solve equation where a $x$ is in the base and in the exponent?

Dylan
  • 16,575
ashold7
  • 167

2 Answers2

0

$$5+5x=4\:c^x\qquad\text{with}\quad c=1.03$$ FIRST PART, analytic calculus (probably this is not the kind of answer expected) :

Let $\quad X=x+1 \quad\to\quad 5X=4\:c^{X-1}\quad\to\quad \frac{5c}{4}X=c^X=e^{X\ln(c)}$

$-X\ln(c)e^{-X\ln(c)}=-\frac{4\ln(c)}{5c} \quad\to\quad -X\ln(c)=W\left(-\frac{4\ln(c)}{5c} \right)$

$W$ is the Lambert's W function. $$x=-1-\frac{1}{\ln(c)}W\left(-\frac{4\ln(c)}{5c} \right)$$ The exact result is : $$x=-1-\frac{1}{\ln(1.03)}W\left(-\frac{4\ln(1.03)}{5.15} \right)\simeq\begin{cases}-0.204829\\184.194\end{cases}$$ The Lambert W function is a multi valuated function. Two roots are obtained in the present case. Both are valable solutions of the equation $\quad 5+5x=4\:(1.03^x)$.

SECOND PART, approximate calculus :

Supposing that we look for a small $x$ only.

$$1.03^x=e^{x\ln(1.03)}\simeq e^{0.03x}\simeq 1+0.03x$$ $$5+5x\simeq 4(1+0.03x)$$ $$x\simeq \frac{1}{-5+0.12}\simeq -0.204918 $$ The accuracy can be increased thanks to further iterative methods (Newton-Raphson for example).

JJacquelin
  • 66,221
  • 3
  • 37
  • 87
  • Thank you for your answer. You seem to know much about Lambert's function. Is it possible to calculate both solutions for the function? Because my calculator does not have the function. – ashold7 Dec 14 '17 at 12:27
  • The LambertW function is implanted in some mathematicam softwares. For example WolframAlpha : http://m.wolframalpha.com/input/?i=-1-1%2Fln%281.03%29LambertW%28-4ln%281.03%29%2F5.15%29 . One can find several papers : http://www.jfepperson.org/2edition-web/lambert.pdf https://www.sciencedirect.com/science/article/pii/S0377042712005213 – JJacquelin Dec 14 '17 at 14:11
  • For the other branche : http://m.wolframalpha.com/input/?i=-1-1%2Fln%281.03%29LambertW%28-1%2C-4ln%281.03%29%2F5.15%29 – JJacquelin Dec 14 '17 at 14:17
0

$$5+5x=4 \cdot 1.03^x \Rightarrow 5(1+x)=1.03^x \cdot 1.03 \cdot {4\over 1.03} \Rightarrow 5(1+x)=1.03^{(1+x)} \cdot {4\over 1.03}$$ $$a=1+x$$ $$5a=1.03^a \cdot {4\over 1.03} \Rightarrow 5 \cdot {1.03\over 4}a=1.03^a \Rightarrow 1.2875a=1.03^a \Rightarrow 1.2875a \cdot 1.03^{-a}=1$$ $$z=e^{ln(z)} \Rightarrow 1.2875a \cdot e^{-a \cdot ln1.03}=1$$ $$y=-a \cdot ln1.03$$ $$1.2875a \cdot {-a \cdot ln1.03\over -a \cdot ln1.03} \cdot e^{-a \cdot ln1.03}=1$$ $$y \cdot e^y=(-a \cdot ln1.03) \cdot e^{-a \cdot ln1.03}={-ln1.03\over 1.2875}$$ Lambert-$W$ function: $y=W(-{ln1.03\over 1.2875})$. $$a=-{1\over ln1.03} \cdot y \Rightarrow 1+x=-{1\over ln1.03} \cdot y$$ $$x=-{1\over ln1.03} \cdot y-1=-{1\over ln1.03} \cdot [W(-{ln1.03\over 1.2875})+ln1.03]$$ Result: $x=-{1\over ln1.03} \cdot [W(-{ln1.03\over 1.2875})+ln1.03]$.

Devid Otladka
  • 34