Number in power of $ log x$ stuck Find $x$ if $$x^{\log 26} - x^{\log 24} = x$$

I can’t fiqure this out , I read all law of log but unfurtunately I can’t solve this. Any help about this?

Sorry for my bad english.
HINT:
$$x^{log(26)}-x^{log(24)}=x$$
$$x^{log(10*2.6)}-x^{log(10*2.4)}=x$$
$$x^{log(10*\frac{13}{5})}-x^{log(10*\frac{12}{5})}=x$$
$$x^{1+log(\frac{13}{5})}-x^{1+log(\frac{12}{5})}=x$$
$$x.x^{log(\frac{13}{5})}-x.x^{log(\frac{12}{5})}-x=0$$
$$x(x^{log(\frac{13}{5})}-x^{log(\frac{12}{5})}-1)=0$$
$$So~,~x=0$$
$$or,x^{log(\frac{13}{5})}-x^{log(\frac{12}{5})}-1=0$$
if you solve this you should get $100$.
HINT:
$$x^{log(26)}-x^{log(24)}=x$$
$$x^{log(10*2.6)}-x^{log(10*2.4)}=x$$
$$x^{log(10*\frac{13}{5})}-x^{log(10*\frac{12}{5})}=x$$
$$x^{1+log(\frac{13}{5})}-x^{1+log(\frac{12}{5})}=x$$
$$x.x^{log(\frac{13}{5})}-x.x^{log(\frac{12}{5})}-x=0$$
$$x(x^{log(\frac{13}{5})}-x^{log(\frac{12}{5})}-1)=0$$
$$So~,~x=0$$
$$or,x^{log(\frac{13}{5})}-x^{log(\frac{12}{5})}-1=0$$
if you solve this you should get $100$.