Question

find the value of x given that 2 log↓10 (2 x - 1) = log↓10 2 + log↓10 (2 x+ 3)​

Answer 1

Answer:

x=5/2,-1/2

Step-by-step explanation:

㏒$(2x-1)^{2}$=㏒2*(2x+3)

=>$(2x-1)^{2}$=2*(2x+3)

$4x^{2}$-4x+1=4x+6

$4x^{2}$-4x-4x+1-6=0

$4x^{2}$-8x-5=0

2x(2x-5)+1(2x-5)=0

x=5/2,-1/2

Answer 2

Answer:

$x = \frac{5}{2}$

Step-by-step explanation:

$2 log_{10}(2x - 1) = log_{10}(2) + log_{10}(2x + 3)$

Since log is defined only for positive values

$2x - 1 > 0 \: and \: 2x + 3 > 0$

$or \: x > \frac{1}{2} \: and \: x > - \frac{3}{2}$

$or \: x > \frac{1}{2}$

Let us simplify the equation

$log_{10}{(2x - 1)}^{2} - log_{10}(2x + 3) = log_{10}(2)$

$log_{10}( \frac{ {(2x - 1)}^{2} }{2x + 3} ) = log_{10}(2)$

$\frac{ {(2x - 1)}^{2} }{2x + 3} = 2$

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

$4 {x}^{2} - 4x + 1 = 4x + 6$

$4 {x}^{2} - 8x - 5 = 0$

$4 {x}^{2} - 10x + 2x - 5 = 0$

$2x(2x - 5) + 1(2x - 5) = 0$

$(2x - 5)(2x + 1) = 0$

$x = \frac{5}{2} or - \frac{1}{2}$

$but \: x > \frac{1}{2}$

$hence \: x = \frac{5}{2} \: is \: the \: solution$