Question

answer this question refer the attachment​

Answer 1

$\large\underline{\sf{Solution-}}$

Given Logarithmic equation is

$\rm :\longmapsto\: log_{2}(8y + 1) - 2 log_{2}(y - 1) = 3 - log_{2}(y + 4)$

can be re-arranged as

$\rm :\longmapsto\: log_{2}(8y + 1) - 2 log_{2}(y - 1) + log_{2}(y + 4) = 3$

We know,

$\boxed{ \tt{ \: xlogy = log {y}^{x} \: }}$

So, using this identity, we get

$\rm :\longmapsto\: log_{2}(8y + 1) - log_{2}(y - 1)^{2} + log_{2}(y + 4) = 3$

We know,

$\boxed{ \tt{ \: logx + logy = log(xy) \: }}$

So, using this, we get

$\rm :\longmapsto\: log_{2}[(8y + 1)(y + 4)] - log_{2}(y - 1)^{2} = 3$

We know,

$\boxed{ \tt{ \: logx - logy = log \frac{x}{y} \: }}$

So, using this, we get

$\rm :\longmapsto\: log_{2}\bigg[\dfrac{(8y + 1)(y + 4)}{ {(y - 1)}^{2} } \bigg] = 3$

We know,

$\boxed{ \tt{ \: log_{x}(y) = z \: \rm \implies\:y = {x}^{z} \: }}$

So, using this, we get

$\rm :\longmapsto\:\dfrac{(8y + 1)(y + 4)}{ {(y - 1)}^{2} } = {2}^{3}$

$\rm :\longmapsto\:(8y + 1)(y + 4) = 8 {(y - 1)}^{2}$

$\rm :\longmapsto\: {8y}^{2} + 32y + 4 + y = 8( {y}^{2} - 2y + 1)$

$\rm :\longmapsto\: {8y}^{2} + 33y + 4 = 8{y}^{2} - 16y +8$

$\rm :\longmapsto\: 33y + 16y= 8 - 4$

$\rm :\longmapsto\: 49y = 4$

$\bf\implies \:\boxed{ \tt{ \: y = \dfrac{4}{49} \: }}$

More to know :-

$\boxed{ \tt{ \: log_{x}(x) = 1 \: }}$

$\boxed{ \tt{ \: log_{ {x}^{y} }( {x}^{z} ) = \frac{z}{y} \: }}$

$\boxed{ \tt{ \: log {e}^{x} = x \: }}$

$\boxed{ \tt{ \: {e}^{logx} = x \: }}$

$\boxed{ \tt{ \: {e}^{y \: logx} = {x}^{y} \: }}$

$\boxed{ \tt{ \: {a}^{ log_{a}(x) } = x \: }}$

$\boxed{ \tt{ \: {a}^{ y \: log_{a}(x) } = {x}^{y} \: }}$