The number of positive integers satisfying x+log10⁡2x+1=xlog10⁡5+log10⁡6 is

x+log10⁡2x+1=xlog10⁡5+log10⁡6 or  xlog10⁡10−log10⁡5+log10⁡2x+1=log10⁡6 or  xlog10⁡2+log10⁡2x+1=log10⁡6 or  log10⁡2x2x+1=log10⁡6 or  22x+2x−6=0 or  2x+32x−2=0∴ 2x=2⇒ x=1