Question

Solve the following equation:
$(1+\frac{1}{2x})\log 3+\log 2=\log (27-\sqrt[x]{x3})$

Answer 1

$(1+\frac{1}{2x})\log 3+\log 2=\log (27-\sqrt[2]{23}){\log 3}^{(1+\frac{1}{2x})}+\log 2=\log (27-\sqrt[x]{x3})2{\left(3\right)}^{(1+\frac{1}{2x})}=27\sqrt[x]{x3}6{\left(\frac{1}{3x}\right)}^{\frac{1}{2}}=27-\left(3^{\frac{1}{x}}\right){(3}^{\frac{1}{x}})+6{\left(3^{\frac{1}{x}}\right)}^{\frac{1}{2}}-27=0(3^{\frac{1}{2x}}+9\left)\right(3^{\frac{1}{2x}}-3)=0$

Therefore, $3^{\frac{1}{2x}}=-9,3$

$3^{\frac{1}{2x}}$ never be $\le 0$

Therefore $3^{\frac{1}{2x}}=3\frac{1}{2x}=1x=\frac{1}{2}$