Question

If log(p+q) (p-q)=-1, then what is the value of log(p+q) (p^2-q^2)?
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Answer 1

Answer:

given that

log (p+q)(p-q) = -1

then log(p+q)(p²-q²) = ?

= log(p+q)(p+q)(p-q)

= log(p+q)²(p-q)

= 2log(p+q)(p-q)

= 2×(-1)

= -2

Answer 2

Answer:

The value of $log(p + q) ( {p}^{2} - {q}^{2} )$ is $log(p + q) - 1$

Step-by-step explanation:

Given,$log(p + q) (p - q) = - 1$

Here we want to find value of $log(p + q) ( {p}^{2} - {q}^{2} )$

We know,

${a}^{2} - {b}^{2} = (a + b)(a - b)$

So,

$log(p + q) ( {p}^{2} - {q}^{2} ) \\ = log(p + q) + log( {p}^{2} - {q}^{2} ) \\ = log(p + q) + log(p + q) (p - q) \\ = log(p + q) - 1$

Some important Logarithm formulas,

$log_{x}(1) = 0 \\ log_{x}(0) = 1 \\ log_{x}(y) = \frac{ log(x) }{ log(y) } \\ log( {x}^{y} ) = y log(x) \\ log(x) + log(y) = log(xy) \\ log(x) - log(y) = log( \frac{x}{y} ) \\ log_{x}(x) = 1$

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