Question

log (3x+2) - log (3x-2) = log5​

Answer 1

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$\bf{ \boxed{ \underline{ \tt{x = 1 \: }}}}}}}$

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$\sf \huge{Question}$

⟹log (3x+2) - log (3x-2) = log5

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step by step explanation:

$\tt{⟹log(3x + 2) - log(3x - 2) = log5}$

using products logarithms

$\tt{⟹log( \frac{3x + 2}{3x - 2}) = log5}$

$\tt{⟹3x + 2 = 5(3x - 2)}$

$\tt{⟹3x + 2 = 15x - 10}$

$\tt{⟹ 15x - 3x = 10 + 2}$

$\tt{⟹12 = 12x}$

$\tt{x = 1}$

I hope it's help uh

Answer 2

Answer :

The value of x is 1

Given :

The equation :

• log(3x + 2) - log(3x - 2) = log5

To Find :

• The value of x

Formula to be used here :

$\star \bf \log( \frac{x}{y} ) = \log(x) - \log(y)$

Solution :

$\sf \log(3x + 2) - log(3x - 2) = log(5) \\ \\ \sf \implies log \{\dfrac{(3x + 2) }{(3x - 2)} \} = log(5) \\ \\ \sf\implies \frac{3x + 2}{3x - 2} = 5 \\ \\ \sf \implies(3x + 2) = 5(3x - 2) \\ \\ \sf \implies3x + 2 = 15x - 10 \\ \\ \sf \implies3x - 15x = - 10 - 2 \\ \\ \sf \implies - 12x = - 12 \\ \\ \sf \implies{x = \frac{ - 12}{ - 12}} \\ \\ \bf \implies x = 1$

Other logarithm formulae :

$\bullet \bf \: \: \log(x.y) = \log(x) + log(y)$

$\bullet \: \: \: \bf \log {a}^{n} = n \log(a)$

$\bullet \: \: \: \bf log_{a {}^{n} }(b {}^{m} ) = \dfrac{m}{n} log_{a}(b)$

$\bullet \: \: \: \bf {a}^{ log_{a}(m) } = m$