# Solve equation with x in the base and x in the exponent

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?

• There is...but not very satisfying many times. Google "Lambert W-function" – DonAntonio Dec 14 '17 at 6:46
• @Arthur: Not really, just change variables to $x+1$ after factoring out the $5$. – Mr. Chip Dec 14 '17 at 7:00
• You will need numerical methods or the Lambert-W-function. The newton-method applying to $f(x)=4\cdot 1.03^x-5x-5$ will probably be best. – Peter Dec 14 '17 at 9:34
• There are $2$ solutions, one near $-0.205$ and the other near $184.194$ – Peter Dec 14 '17 at 9:40

$$5+5x=4\:c^x\qquad\text{with}\quad c=1.03$$ 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).

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