How can I solve this type of equation $2x=3^x+2$. I tried taking the logarithm of both sides but it doesn't solve $x$. I also tried to search it on the internet but I don't know what to search.
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Are you sure it has a solution. Just by eyeballing, we can say x cannot be negative or zero. And when x is positive, RHS is possibly always bigger than LHS. – Juanito Sep 26 '14 at 06:13
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Observe that $3^x$ is always positive the right side of the equality is strictly greater than 2. The left side takes this form when $x>1$. So we need only look for a solution in this area. Now we can approximate the solution for this as follows.
Try $x=1$. Then $2(1) =2$ and $3^1+2 =5$.
Since the exponential increases faster than the left, linear side, this equation has no solutions.
Eoin
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As already answered by Eoin, there is no solution in the real domain. However, this equation has a solution which can be expressed using the Lambert function; this solution is $$x=1-\frac{W\left(-\frac{3 \log (3)}{2}\right)}{\log (3)} \approx 0.968921-1.44932 i$$
In a more general manner, the equation $$a x=b^x+c$$ has a solution which is $$x=\frac{c}{a}-\frac{W\left(-\frac{\log (b) b^{\frac{c}{a}}}{a}\right)}{\log (b)}$$ which would be real if $$\frac{\log (b) b^{\frac{c}{a}}}{a}<\frac{1}{e}$$ For example, using $a=10,b=3,c=8$ would lead to two solutions $$x_1=\frac{4 \log (3)-5 W\left(-\frac{1}{10} 3^{4/5} \log (3)\right)}{5 \log (3)} \approx 1.15614$$ $$x_2=\frac{4 \log (3)-5 W_{-1}\left(-\frac{1}{10} 3^{4/5} \log (3)\right)}{5 \log (3)} \approx 2.66152$$
Claude Leibovici
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