Question

log (6x^2-5x+1) base 1-2x - log (4x^2 -4x+1) base 1-3x = 2​

Answer 1

$\textbf{Given:}$

$log_{1-2x}(6x^2-5x+1)-log_{1-3x}(4x^2-4x+1)=2$

$\textbf{To find:}$

$\text{The values of x}$

$\textbf{Solution:}$

$log_{1-2x}(6x^2-5x+1)-log_{1-3x}(4x^2-4x+1)=2$

$log_{1-2x}(3x-1)(2x-1)-log_{1-3x}(1-2x)^2=2$

$log_{1-2x}(1-3x)(1-2x)-2\,log_{1-3x}(1-2x)=2$

$log_{1-2x}(1-3x)+log_{1-2x}(1-2x)-2\,log_{1-3x}(1-2x)=2$

$log_{1-2x}(1-3x)+1-2\,log_{1-3x}(1-2x)=2$

$log_{1-2x}(1-3x)-2\,log_{1-3x}(1-2x)=1$

$log_{1-2x}(1-3x)-\dfrac{2}{log_{1-2x}(1-3x)}=1$

$t-\dfrac{2}{t}=1$ $\text{where t=log_{1-2x}(1-3x) }$

$\dfrac{t^2-2}{t}=1$

$t^2-t-2=0$

$(t-2)(t+1)=0$

$\implies\,t=2,-1$

$\textbf{when t=2}$

$log_{1-2x}(1-3x)=2$

$\implies\,(1-2x)^2=1-3x$

$1+4x^2-4x=1-3x$

$4x^2-x=0$

$x(4x-1)=0$

$x=0,\frac{1}{4}$

$\textbf{when t=-1}$

$log_{1-2x}(1-3x)=-1$

$\implies\,(1-2x)^-1=1-3x$

$\dfrac{1}{1-2x}=1-3x$

$(1-2x)(1-3x)=1$

$1-5x+6x^2=1$

$6x^2-5x=0$

$x(6x-5)=0$

$x=0,\frac{5}{6}$

$\text{But x=0 makes base of the logarithm as 1}$

$\therefore\textbf{The values of x are \bf\frac{1}{4} \;\textbf{and}\; \bf\frac{5}{6} }$

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