Question

logx (8x - 3) - logx 4 = 2.​

Answer 1

Answer:

................... ..........

Answer 2

Given:-

• $\sf log_{x}(8x - 3) - log_{x}(4) = 2$

To find:-

• The value of x

Answer:-

Given that,

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

We know that,

$\red{ \bigstar} \boxed{ \sf{ log_{b}(a) - log_{b}(c) = log_{b} \bigg( \dfrac{a}{c} \bigg) }}$

So, we can write the given equation as,

$\sf \longrightarrow log_{x} \bigg \{ \dfrac{ (8x - 3)}{4} \bigg \} = 2$

We know that,

$\blue{ \bigstar} \boxed{ \sf{ if \: log_{b}(a) = c, \: then \: a = {b}^{c} }}$

(this is know as exponential form)

So, writing in exponential form,

$\sf \longrightarrow \dfrac{8x - 3}{4} = {x}^{2}$

$\sf \longrightarrow 8x - 3 = 4 {x}^{2}$

$\sf \longrightarrow 4 {x}^{2} - 8x + 3 = 0$

On splitting the middle term,

$\sf \longrightarrow 4 {x}^{2} - 2x - 6x + 3 = 0$

$\sf \longrightarrow 2x(2x - 1) - 3(2x - 1) = 0$

$\sf \longrightarrow (2x - 1)(2x - 3)= 0$

So,

If (2x - 1) = 0,

$\sf 2x - 1 = 0$

$\longrightarrow \underline{ \underline{ \sf{ \green{ x_{1} = \dfrac{1}{2} }}}}$

If (2x - 3) = 0,

$\sf 2x - 3 = 0$

$\longrightarrow \underline{ \underline{ \sf{ \green{ x_{2} = \dfrac{3}{2} }}}}$

(Both of these solutions follows the conditions necessary for logarithm).