Simple looking log problem How would I solve this for $x$?
The original problem is
$$x+x^{\log_{2}3}=x^{\log_{2}5}$$
I have tried to reduce it down to this,
$$x^{\log_{10}3}+x^{\log_{10}2}=x^{\log_{10}5}$$
I have been stuck on this for a while now, can't seem to figure out the trick to this.
Sorry if I tagged this wrong, not sure what tags it should be.
 A: Set $y=x^{1/\log 2}$, then your problem becomes $$1+y^{\log 3}=y^{\log 5}$$
There's no real hope of making this a polynomial, because $\log 3$ and $\log 5$ are linearly independent over $\mathbb{Q}$.  Hence the best you can do is solve it numerically, which wolfram does as $$y\approx 5.33532$$
and then $$x\approx 1.65538$$
A: Oops!
My answer was wrong,
because I wrote the first term of the equation as
$1$
instead of $x$.
I will correct this now,
and hope I do not make any other errors.

The equation is
$x+x^{\log_{2}3}=x^{\log_{2}5}$.
Eliminating the obvious root of
$x=0$,
this becomes
$1+x^{\log_{2}3/2}=x^{\log_{2}5/2}$
or
$1=x^{\log_{2}5/2}-x^{\log_{2}3/2}
$
For any $v$,
$x^{\log_2 v}
= (2^{\log_2 x})^{\log_2 v}
= 2^{(\log_2 x)(\log_2 v)}
= 2^{(\log_2 v)(\log_2 x)}
= (2^{\log_2 v})^{\log_2 x}
= v^{\log_2 x}
$,
so the equation becomes
$1=(5/2)^{\log_2 x}-(3/2)^{\log_2 x}
$
Letting $y = \log_2 x$,
this is
$1 = (5/2)^y-(3/2)^y
$.
Let $f(y) = (5/2)^y-(3/2)^y-1$.
If $y < 0$, letting $z = -y$,
$f(y)=(2/5)^z-(2/3)^z-1
< 0
$
for $z > 0$
since $(2/5)^z<(2/3)^z$.
$f(0) = -1$,
$f(1) = 0$
(aha! a root at $x=2$),
and
$f(2) = (25/4)-(9/4)-1 = 3$.
$\begin{align}
f'(y) 
&= (\ln (5/2))(5/2)^y-(\ln (3/2))(3/2)^y\\
&>(\ln (3/2))(5/2)^y-(\ln (3/2))(3/2)^y\\
&=(\ln (3/2))((5/2)^y-(3/2)^y)\\
&=(\ln (3/2))(3/2)^y((5/3)^y-1)\\
&> 0 \text{ for } y>0\\
\end{align}
$
so the only root is at $y=1$,
so the only roots of the original equation are
$x=0$ and $x=2$.

Here is my original answer:
First,
the base does not matter,
since only the ratio of logs is used.
Second,
the basic law of logs is that
if $u = \log_v w$,
then $w = v^u = v^{\log_v w}$.
From this
we can deduce that
$\dfrac{\log_c a}{\log_c b}
= \log_b a$.
To see this,
if $u = \log_c a$ and $v = \log_c b$,
then
$c^u = a$ and $c^v = b$
so
$c = b^{1/v}$
and $a = c^u = (b^{1/v})^u
= b^{u/v}$,
so
$u/v = \log^b a$.
Therefore
$x = 2^{\log_2 x}$,
so
$x^{\log_2 3}
=(2^{\log_2 x})^{\log_2 3}
=2^{(\log_2 x)(\log_2 3)}
=2^{(\log_2 3)(\log_2 x)}
=(2^{\log_2 3})^{\log_2 x}
=3^{\log_2 x}
$
and, similarly
$x^{\log_2 5}
=5^{\log_2 x}
$.
The equation now becomes
$1 + 3^{\log_2 x} = 5^{\log_2 x}$.
Let $y = \log_2 x$.
Then
$1 + 3^y = 5^y$.
As vadim123 stated,
this can only be solved numerically.
Once $y$ has been gotten,
$x = 2^y$.
To show that this is the only root,
let $f(y) = 1+3^y-5^y$.
$f(0) = 1$,
$f(1) = -1$,
and
$f'(y) = (\ln 3)3^y - (\ln 5)5^y
< 0$
 for $y \ge 0$,
so the root between $0$ and $1$
is the only one.
A: If you are sure this is true;
$$
x^{log_{10} 3} + x^{log_{10} 2} =x^{log_{10} 5}
$$
then you can use use inverse function of exponential function property
$$
f(f^{-1}(x)) = b^{lob_{b} x} = x;
$$
Then your problem comes down to;
$$
3^{log_{10} x} + 2^{log_{10} x} =5^{log_{10} x}
$$
where x = 10.
A: I am not sure how you arrived at the reduced equation. Let's solve the original equation. Let $x=2^{y+1}$. Then
\begin{align*}
&x+x^{\log_{2}3}=x^{\log_{2}5}\tag{1}\\
\Leftrightarrow&2(2^y)+3(3^y)=5(5^y)\\
\Leftrightarrow&2(5^y-2^y) + 3(5^y-3^y) = 0.\tag{2}
\end{align*}
Since the functions $f(y)=5^y-2^y$ and $g(y)=5^y-3^y$ are stictly increasing in $y$ (exercise: show that $(5^b-2^b)-(5^a-2^a)>0$ whenever $b>a$) and $f(0)=g(0)=0$, the only positive solution of $(2)$ occurs at $y=0$, i.e. $x=2$. As $x=0$ is also a solution, $x=0,2$ are the only nonnegative solutions of $(1)$.
