Solving an equation with a variable as both a part of a term and exponent

Question

The problem : $$ 60 + 10x=2^x$$

I've tried using Logarithms which leads to :

$$\log(60+10x)=x\log 2$$

The $\log(60+x)$ is the tricky part

Answer 1

Where in the world did you get this problem? There is no "elementary" solution to such an equation.

What you can do (if you really need to write "x= something") is write $2^x$ as $e^{ln(2^x)}= e^{xln(2)}$ so that the equation becomes $60+ 10x= e^{xln(2)}$.

Now let u= 60+ 10x so that $x= \frac{u- 60}{10}= \frac{u}{10}- 6$ and we have $u= e^{\left(\frac{u}{10}- 6\right)ln(2)}= e^{\frac{ln(2)}{10}u}e^{-6ln(2)}$.

$ue^{-\frac{ln(2)}{10}u}= e^{-6ln(2)}$.

Now let $v= -\frac{ln(2)}{10}u$ so $u= -\frac{10}{ln(2)}v$ and the equation becomes $-\frac{10}{ln(2)}ve^v= e^{-6ln(2)}$.

$ve^v= \frac{ln(2)e^{-6ln(2)}}{10}$

Now take the "Lambert's W function" (which is defined as the inverse function to $f(x)= xe^x$) https://en.wikipedia.org/wiki/Lambert_W_function of both sides to get

$v= W\left(-\frac{ln(2)}{10}e^{6ln(2)}\right)$!

We need to work back to x. Since $v= -\frac{ln(2)}{10}u$ $u= -\frac{10}{ln(2)}v= -\frac{10}{ln(2)}W\left(-\frac{ln(2)}{10}e^{6ln(2)}\right)$

And since u= 60+ 10x, Finally $x= \frac{u- 60}{10}= \frac{u}{10}- 6= \frac{1}{ln(2)}W\left(-\frac{ln(2)}{10}e^{6ln(2)}\right)- 6$!