How would I solve this exponential = linear function?

Question

Would I solve for $ab^x = kx+m$ ? I am lost to how to solve it

This equation is what I need help on: $7\cdot 8^x +3 = 9x+2$

Im puzzled to how to solve this.

Answer 1

$\require{begingroup} \begingroup$ $\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

\begin{align} ab^x &= kx+m \tag{1}\label{1} \end{align}

There is absolutely no need for any numerical methods in this case, since the Lambert $\W$ function was specifically invented to express a solution(s) to such kind of equations, as the symbol $\sqrt{\ }$ was invented to represent a solution $x=\pm\sqrt a$ for equation $x^2=a$, and the most suitable numerical methods were well-studied already, and were implemented in software.

The equation \eqref{1} just needs to be transformed to the form $u\exp(u)=v$ to apply $\W(u\exp(u))=\W(v)$ to find $u=\W(v)$.

\begin{align} \frac ak\,\exp(x\ln(b))\,\ln(b) &= x\,\ln(b)+\frac mk\,\ln(b) \tag{2}\label{2} ,\\ \frac ak\,\exp\left(x\ln(b)+\frac mk\,\ln(b)-\frac mk\,\ln(b)\right)\,\ln(b) &= x\,\ln(b)+\frac mk\,\ln(b) \tag{3}\label{3} ,\\ \frac ak\,\exp\left(x\ln(b)+\frac mk\,\ln(b)\right)\exp\left(-\frac mk\,\ln(b)\right)\,\ln(b) &= x\,\ln(b)+\frac mk\,\ln(b) \tag{4}\label{4} . \end{align}

Let $\left(x+\frac mk\right)\,\ln(b)=y$ for now, so \eqref{4} becomes

\begin{align} \frac ak\,\exp(y)\exp\left(-\frac mk\,\ln(b)\right)\,\ln(b) &= y \tag{5}\label{5} ,\\ \frac ak\,\exp\left(-\frac mk\,\ln(b)\right)\,\ln(b) &= y\exp(-y) \tag{6}\label{6} , \end{align}

\begin{align} -y\exp(-y) &= -\frac ak\,\exp\left(-\frac mk\,\ln(b)\right)\,\ln(b) \tag{7}\label{7} ,\\ -y\exp(-y) &= -\frac{a\,\ln(b)}{k\,\exp\left(\frac mk\,\ln(b)\right)} = -\frac{a\,\ln(b)}{k\,b^{m/k}} \tag{8}\label{8} ,\\ \W(-y\exp(-y)) &= \W\left( -\frac{a\,\ln(b)}{k\,b^{m/k}} \right) \tag{9}\label{9} ,\\ \tag{10}\label{10} -y &= \W\left( -\frac{a\,\ln(b)}{k\,b^{m/k}} \right) ,\\ \left(x+\frac mk\right)\,\ln(b) &= -\W\left( -\frac{a\,\ln(b)}{k\,b^{m/k}} \right) \tag{11}\label{11} ,\\ x &= - \frac1{\ln(b)} \W\left( -\frac{a\,\ln(b)}{k\,b^{m/k}} \right) -\frac mk \tag{12}\label{12} . \end{align}

At this point we have the solution for $x$ in terms of the Lambert $\W$ function.

The last and a very important step is to find out, is there any real solutions, and how many of them. And there is a well-known condition, based on the value of the operand of the $\W(z)$ function, in this case it is $z=-\frac{a\,\ln(b)}{k\,b^{m/k}}$, so

\begin{align} \begin{cases} z<-\tfrac1\e &\Rightarrow \text{no real solutions} ,\\ z\ge0 &\Rightarrow \text{one real solution}, x = -\frac1{\ln(b)}\Wp(z)-\frac mk ,\\ -\tfrac1\e < z<0 &\Rightarrow \text{two real solutions}, x_0=-\frac1{\ln(b)}\Wp(z)-\frac mk ,\quad x_1=-\frac1{\ln(b)}\Wm(z)-\frac mk ,\\ z=-\tfrac1\e &\Rightarrow \text{in this case} \Wp(-\tfrac1\e)=\Wm(-\tfrac1\e)=-1, \text{ so } x=\frac1{\ln(b)}-\frac mk . \end{cases} \end{align}

For $a=7,\, b=8,\,k=9,\, m=-1$ we have the argument of $\W(z)$

\begin{align} z&=-\frac{a\,\ln(b)}{k\,b^{m/k}} =-\frac79\,\ln(8)\,8^{1/9}\approx -2.03772502141 <-\tfrac1\e , \end{align}

so indeed, in this case there are no real solutions.

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Answer 2

Yes indeed, plotting the linear function and the exponential function (or making a spreadsheet table) will show you clearly that there is no point of intersection. And if there were a point of intersection for such a pair of functions (with different coefficients), then you would generally need to solve for it numerically.

Ob the other hand, one can take the derivatives of both functions:

d/dx (7 * 8^x + 3) = 7*ln(8)*8^x

d/dx (9x +2) = 9

The exponential function has a slope that is always positive (in fact, the second derivative is also positive which means its curvature is always upward).

It can be seen that the exponential function has a horizontal asymptote of 3 (as x approaches negative infinity), where the slope is extremely small (but positive). As the curve passes the y-axis, the slope equals 14.56 (=7*ln(8)) and then continues to grow (exponentially). By setting the two derivatives equal to each other (i.e., =9) and solving, we can find the point at which the curves are the closest (vertically):

x= ( ln(9)-ln[7*ln8] ) / ln(8) = -0.2312

We can plug that x value into each equation and see that the exponential curve is 7.408963 above the linear equation. At that point, the exponential had been inching closer to the straight line (its slope to the left was less than 9), but after this point (moving to the right), the exponential curve will grow faster than a slope of 9 and will climb away from the straight line.