Question

solve for x:- logx(8x-3)-logx4=2​

Answer 1

Solution :

$\displaystyle \mathrm{log_{x}(8x-3)-log_{x}4=2}$

$\displaystyle \implies \mathrm{log_{x}\frac{8x-3}{4}=2}$

Taking powers to x, we get

$\displaystyle \implies \mathrm{x^{(log_{x}\frac{8x-3}{4})}=x^{2}}$

$\displaystyle\implies \mathrm{\frac{8x-3}{4}=x^{2}}$

$\displaystyle\implies \mathrm{4x^{2}=8x-3}$

$\displaystyle\implies \mathrm{4x^{2}-8x+3=0}$

$\displaystyle\implies \mathrm{4x^{2}-6x-2x+3=0}$

$\displaystyle\implies \mathrm{2x(2x-3)-1(2x-3)=0}$

$\displaystyle\implies \mathrm{(2x-3)(2x-1)=0}$

Either $\displaystyle \mathrm{2x - 3 = 0}$ or, $\displaystyle \mathrm{2x - 1 = 0}$

$\displaystyle \implies\boxed{\mathrm{x = \frac{3}{2}\:\,\:,\:\,\:\frac{1}{2}}}$

which be the required solution

Rule :

$\displaystyle \mathrm{a^{(log_{a}b)}=b}$