Question

DIU
rii) log(3x + 2) + log(3x - 2) = 5 log 2.​

Answer 1

=> log(3x + 2) + log(3x - 2) = 5 log 2

=> log(3x + 2)(3x - 2) = 5 log 2

=> log(3x + 2)(3x - 2) = log 2⁵

=> log(9x²- 4) = log 2⁵

Taking anti log both sides :-

=> (9x²- 4) = 2⁵

=> 9x²- 4 = 32

=> 9x² = 32 +4

=> 9x² = 36

=> x² = (36)/9

=> x² = 4

=> x = √4

=> x = ±2

but x cannot be negative.

Therefore , x = 2

Answer : 2

Additional Information :

$\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\ ^{a} log \ a= 1\\\\\circ \ ^{a}log \ 1 = 0 \\\\\circ \ ^{a ^{n}} log \ b^{m}= \dfrac{m}{n} \times\:^{a}log \ b \\\\\circ \ ^{a^{m}} \ log \ b^{m} = \ ^{a}log \ b \\\\\circ \ ^{a}log \ b = \dfrac{1}{^{b}log \ a} \\\\\circ \ ^{a}log \ b = \dfrac{^{m}log \ b}{^{m} log \ a} \\\\\circ \ a^{^{a} logb} = b \\\\\circ \ ^{a}log \ b + ^{a}log \ c = \ ^{a}log(bc) \\\\\circ \ ^{a}log \ b -\: ^{a}log \ c = \ ^{a}log \left( \dfrac{b}{c} \right) \\\circ \ ^{a}log \ b \:\cdot\: ^{a}log \ c = \ ^{a}log \ c \\\\\circ \ ^{a}log \left( \dfrac{b}{c} \right) = \ ^{a}log \left(\dfrac{c}{b}\right)\end{minipage}}}$

Answer 2

Answer:

★Question★

• Solve for x :- log(3x + 2) + log(3x - 2) = 5 log 2.

★Answer★

★Given :-

• log(3x + 2) + log(3x - 2) = 5 log 2.

★To find :-

• The value of x.

★Formula used:-

$\bf \: loga + logb = logab$

$\bf \: log( {x}^{y} ) = y log(x)$

$\bf \: {x}^{2} - {y}^{2} = (x + y)(x - y)$

$\bf \: log(x) \: is \: defined \: when \: x > 0$

★Solution :-

$\bf \:log(3x + 2) + log(3x - 2) = 5 log 2.$

$\bf\implies \: log((3x + 2)(3x - 2)) = log( {2}^{5} )$

★ On comparing, we get

$\bf\implies \: {9x}^{2} - 4 = 32$

$\bf\implies \:9 {x}^{2} = 32 + 4$

$\bf\implies \:9 {x}^{2} = 36$

$\bf\implies \: {x}^{2} = 4$

$\bf\implies \:x = 2$