${log}_{11} (2 x - 1) = 1 - {log}_{11} (x + 4)$ $log_11(2x-1)=1-log_11(x+4)$ What is $x$ $x$ ?

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${log}_{11} (2 x - 1) = 1 - {log}_{11} (x + 4)$ $log_11(2x-1)=1-log_11(x+4)$ What is $x$ $x$ ?

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Answer 1 · 2017-12-17T13:40:03

${log}_{11} (2 x - 1) = 1 - {log}_{11} (x + 4)$ $log_11(2x-1)=1-log_11(x+4)$

$\Rightarrow {log}_{11} (2 x - 1) + {log}_{11} (x + 4) = 1$ $=>log_11(2x-1)+log_11(x+4)=1$

$\Rightarrow {log}_{11} [(2 x - 1) (x + 4)] = {log}_{11} 11$ $=>log_11[(2x-1)(x+4)]=log_11 11$
$\Rightarrow (2 x - 1) (x + 4)] = 11$ $=>(2x-1)(x+4)]= 11$

$\Rightarrow 2 x^{2} - x + 8 x - 4 - 11 = 0$ $=>2x^2-x+8x-4- 11=0$

$\Rightarrow 2 x^{2} + 7 x - 15 = 0$ $=>2x^2+7x-15=0$

$\Rightarrow 2 x^{2} + 10 x - 3 x - 15 = 0$ $=>2x^2+10x-3x-15=0$

$\Rightarrow 2 x (x + 5) - 3 (x + 5) = 0$ $=>2x(x+5)-3(x+5)=0$

$\Rightarrow (x + 5) (2 x - 3) = 0$ $=>(x+5)(2x-3)=0$

So $x = - 5 and x = \frac{3}{2}$ $x=-5 and x=3/2$

But $x = - 5$ $x=-5$ is not valid for the given equation ${log}_{11} (2 x - 1) = 1 - {log}_{11} (x + 4)$ $log_11(2x-1)=1-log_11(x+4)$

Hence $x = \frac{3}{2}$ $x=3/2$

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${log}_{11} (2 x - 1) = 1 - {log}_{11} (x + 4)$ $log_11(2x-1)=1-log_11(x+4)$ What is $x$ $x$ ?

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${log}_{11} (2 x - 1) = 1 - {log}_{11} (x + 4)$ $log_11(2x-1)=1-log_11(x+4)$ What is $x$ $x$ ?