# number of real solution of exponential and logarithmic equation

The number of solution of the equation $(x-2)+2\log_{2}(2^x+3x)=2^x$

$\bf{My\; Try::}$ We can write it as $2\log_{2}(2^x+3x) = 2^x-x+2$

Now Let $f(x) = 2\log_{2}(2^x+3x)$ and $g(x)=2^x-x+2$

Here we have to calculate number of points where $f(x)$ and $g(x)$ intersect each other

So here we will find nature of $f(x)$ and $g(x)$

So $$f'(x) = \frac{1}{\ln_{e}(2)}\frac{2\cdot (2^x\ln (2)+3)}{2^x+3x}>0\forall x \in \bf{Domian}$$ bcz $2^x+3x>0$ for logarithmic function.

and $f(0) = 0$. So function $f(x)$ is strictly Increasing function.

Now $$g'(x) = 2^x\ln 2-1 >0\forall x \geq 1$$

and for $0\leq x<1\;,$ Then $1\leq 2^x<2\Rightarrow 1\cdot \ln(2)-1 \leq 2^x\ln(2)-1<2\cdot \ln(2)-1$

So minimum of function at $x=\alpha \in (0,1)$.

Now $g'(x)>0$ for $x>\alpha \in (0,1)$ and $g'(x)<0$ for $x=\alpha \in (0,1)$

So solution of $f(x) = g(x)$ exists when $x>0$

Now how can i solve it after that, Help required, Thanks

Plot the function $f(x)=(x-2)+2\log_{2}(2^x+3x)-2^x$,you can find there are only two solutions.
So it becomes 1. to show $f(2)>0$ and $f(\pm \infty) <0$. 2. it is concave (i.e. second derivative is negative.