How do I solve $\log_2 (3x+2) = \frac{(2^x)-2}{3}$? Does it need to be done using graphs?

Question

$$\log_2 (3x+2) = \frac{(2^x)-2}{3}$$ Finding intersection of both functions provided that one is the inverse of another.

Answer 1

By using the Lambert-W function we can calculate a closed form solution. For solutions, we have to equivalently solve (see my comment) the equation $$x=\frac{2^x-2}3$$ $$3\left(x+\frac23\right)=2^x$$ $$3\cdot2^{2/3}\left(x+\frac23\right)=2^{x+2/3}$$ $$3\cdot2^{2/3}\left(x+\frac23\right)=e^{\ln{(2)}(x+2/3)}$$ $$\left(x+\frac23\right)e^{-\ln{(2)}(x+2/3)}=\frac1{3 \cdot 2^{2/3}}$$ $$-\ln{(2)}\left(x+\frac23\right)e^{-\ln{(2)}(x+2/3)}=-\frac{\ln{(2)}}{3 \cdot 2^{2/3}}$$ $$-\ln{(2)}\left(x+\frac23\right)=W_k\left(-\frac{\ln{(2)}}{3 \cdot 2^{2/3}}\right)$$ $$x+\frac23=-\frac{W_k\left(-\frac{\ln{(2)}}{3 \cdot 2^{2/3}}\right)}{\ln{(2)}}$$ $$\therefore x=-\frac{W_k\left(-\frac{\ln{(2)}}{3 \cdot 2^{2/3}}\right)}{\ln{(2)}}-\frac23$$ Where the two real solutions are given by the branches $k=-1$ and $k=0$ of the Lambert-W function i.e. $$x=-\frac{W_{-1}\left(-\frac{\ln{(2)}}{3 \cdot 2^{2/3}}\right)}{\ln{(2)}}-\frac23\approx3.717149949436548429359099694504744579838\dots$$ $$x=-\frac{W_0\left(-\frac{\ln{(2)}}{3 \cdot 2^{2/3}}\right)}{\ln{(2)}}-\frac23\approx-0.4170078734328476800738659669535523944241\dots$$

Answer 2

There is no closed form algebraic solution. You will need graphical or numerical methods, as in the picture.

Answer 3

According to the fact that the equal functions are inverse to each other, we can simplify the equation from $$\log_2 (3x+2) = \frac{(2^x)-2}{3}$$ to $$\log_2 (3x+2) = x$$ or according to the definition of logarithms ($\log_a b = c \Leftrightarrow b = a^c$) to $$3x+2=2^x$$ Now, write this last equation a little bit differently as $$3x+2-2^x=0$$ and define a function $$f(x)=3x+2-2^x$$ The main goal now is to make this function equal to zero or $f(x)=0$. This is an equivalent task to the very first equation above. Just by looking at this function $f(x)$ we can make an educated guess and choose $x=4$. This choice of $x=4$ makes $f(x)=-2<0$. Not yet $0$. But $x_0 = 4$ is just an initial guess. Then we use the recursion formula (newton formula / newton method) $$x_{i+1}=x_{i}-\frac{f(x_i)}{f'(x_{i})}$$ to get to better results for $x$ that make $f(x)=0$, because $x_0 = 4$ apparently doesn't make $f(x)=0$. To use this formula we need to find the derivative $f'(x)$ of $f(x)$. This is: $$f'(x) = 3 - \ln(2) \cdot 2^x$$ Then we are able to find $$x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = 4 - \frac{3\cdot 4+2-2^4}{3 - \ln(2) \cdot 2^4} \approx 3.753 $$ This result for $x_1 \approx 3.753$ makes the function $f(x_1) \approx -0.223$, which still isn't $0$. Once more we can find another $$x_2 = x_1 - \frac{f(x_1)}{f'(x_1)} = 3.753 - \frac{3\cdot 3.753 +2-2^{3.753} }{3 - \ln(2) \cdot 2^{3.753}} \approx 3.718 $$ This result for $x_2 \approx 3.718$ makes the function $f(x_2) \approx -0.0052$, which still isn't $0$. but we're getting closer. Now calculate $$x_3 = x_2 - \frac{f(x_2)}{f'(x_2)} = 3.718 - \frac{3\cdot 3.718 +2-2^{3.718} }{3 - \ln(2) \cdot 2^{3.718}} \approx 3.717$$ Now $f(x_3)=f(3.717)\approx 0.0009$, so it gets pretty closer to zero as we move to $x_4, x_5, \cdots $ and so on.

If we would have chosen another initial guess near to the other solutions e.g. $y_0 = -1$, then we would have had $f(y_0) = -1.5$. We then get to better solutions using the very same newton formula like:

$$y_1 = y_0 - \frac{f(y_0)}{f'(y_0)} = -1 - \frac{3\cdot (-1) +2-2^{-1} }{3 - \ln(2) \cdot 2^{-1}} = -1 - \frac{-1.5 }{3 - \ln(2) \cdot 2^{-1}} \approx -0.435$$

$$y_2 = y_1 - \frac{f(y_1)}{f'(y_1)} \approx -0.435 - \frac{3\cdot (-0.435) +2-2^{-0.435} }{3 - \ln(2) \cdot 2^{-0.435}} = -1 - \frac{-0.045 }{3 - \ln(2) \cdot 2^{-0.435}} \approx -0.417$$

So $y_2 \approx -0.417$ makes our function $f(x = y_2) \approx -0.045$, which is practically $0$.

The two solutions were found numerically to be approximately $3.717$ and $-0.417$.