Please help me to solve this equation $3x+3^{\ln{x}}-4=0$ Please help me to solve this equation 
$$3x+3^{\ln{x}}-4=0$$
I found $x = 1$ in a plot W.A, but can I find x by Algebraic solution?
Thanks
 A: $$ 3x-3=1-3^{\ln{x}}$$
Either  $$3x - 3 \ge 0 \text{ and } 1-3^{\ln{x}}\ge0$$
         $$ 3x \ge 3 \text{ and } 1\ge 3^{\ln{x}}$$
$$ x \ge 1 \text{ and } 1 \ge x$$ 
$$ \text{Then } x = 1$$
OR
$$3x - 3 \le 0 \text{ and } 1 - 3^{\ln{x}} \le 0$$
         $$ 3x \le 3 \text{ and } 1 \le 3^{\ln{x}}$$
$$ x \le 1 \text{ and } 1 \le x$$
$$ \text{Then } x = 1$$
A: $\ln x$ is defined only if $x>0$.
Also $3^{\ln{x}}=4-3x \Longrightarrow 4-3x>0 \Longrightarrow x<\frac{4}{3}$.
It means that $0<x<\frac{4}{3}$.
And $x=1$ is just lucky guess from this interval. It's natural first choice because then $\ln{x}=0$ and $3^{\ln{x}}=3^0=1$. And it's only solution because $3^{\ln{x}}$ is strictly increasing and $4-3x$ is strictly decreasing.
There is no normal algebraical way because such exercises are constructed to be solved with such "lucky guess method". In general case there isn't a nice solution.
For example, $3^{\ln{x}}=5-3x$ has "ugly" irrational solution $x=1.24325...$ which can be found only by numerical methods.

I thought that Ahmed with his answer showed that I am wrong. Sadly it wasn't true. His algebraical solution still involves lucky guess.
It works in such way.
You have equation $f\left(x\right)=g\left(x\right)$ with only solution $x_0$ which gives $f\left(x\right)=g\left(x\right)=y_0$.
From it follows $f\left(x\right)-y_{0}=g\left(x\right)-y_{0}$.
Then you prove that $f\left(x\right)-y_{0}$ and $g\left(x\right)-y_{0}$ have equal signs only when $x=x_0$. (When both values are $0$.)
TL;DR; You still have to guess. In this case it's $y_0$.
A: Whoever created this problem was having some fun.
Note that, for any positive real $a$,
$x^a = 1$ for $x = 1$.
So, for any positive real $b$,
$f(x) = x^a + b x - (b+1)$
has $x = 1$ as a root.
Also,
if $a > 1$
(which is true for $a = \ln 3$), 
$f'(x) = a x^{a-1} + b$
is positive for $x > 0$,
so $f$ can have at most one positive real root.
You can easily make the problem
look more esoteric
by using
$x^{\pi} + x \log_2{e} - \log_2{(2e)}$
which, by the same reasoning,
has $1$ as its only positive real root.
