A beautiful exponential equation

 

                     Topic 2. Find the real numbers $x$ for which

$3^x+3^{[x]}+3^{\{x\}}=4 \tag{E}$

                 ($[x]$ and $\{x\}$ represent the integer part and the fractional part of the real

                  number $x$, respectively)

ANSWER CiP

$$x_1=1-log_32,\;x_2=log_3{11}-2log_32-1$$

                             Solution CiP

              Because $x=[x]+\{x\}$ the equation is written equivalently

$3^{[x]} \cdot 3^{\{x\}}+3^{[x]}+3^{\{x\}}=4\;\;\;|+1\;\Leftrightarrow$

$\Leftrightarrow\;(3^{[x]}+1)\cdot (3^{\{x\}}+1)=5.$

          We obtain from the above  $3^{\{x\}}+1=\frac{5}{3^{[x]}+1}$  and because \frac $0\leqslant \{x\}<1$, therefore  $2\leqslant 3^{\{x\}}+1<4$, we must have

$$2\leqslant \frac{5}{3^{[x]}+1}<4.$$

The last inequalities imply

  $\frac{1}{4}<3^{[x]}\leqslant \frac{3}{2}$ 

and because  $[x]\in \{\dots,-2,\;-1,\;0,\;1,\;\dots\}$, it results $[x]=0$ or $[x]=-1$.

          If $[x]=0$, then $\{x\}=x$, and from (E) we obtain $3^x+1+3^x=4$ so $3^x=\frac{3}{2}$, hence $x=log_3\left ({\frac{3}{2}}\right )=log_33-log_32=1-log_32.$

If  $[x]=-1$, then $\{x\}=x+1$, and from (E) we obtain $3^x+\frac{1}{3}+3\cdot 3^x=4$ so $3^x=\frac{11}{12}$, hence $x=log_3 \left (\frac{11}{12} \right )=log_3{11}-log_3{(2^2\cdot 3)}=log_3{11}-2log_32-1$.

     Both values ​​found for $x$ verify equation (E).

$\blacksquare$

                         Official Solution

               It is shown that if  $x\geqslant 1$ or if  $x<-1$ then there are NO solutions.

It remains to study the cases  $x\in [0,1)$, when $[x]=0$, and  $x\in [-1,0)$, when  $[x]=-1$, and find, as with us, the answer.