Question

If log(a+b)(a-b) = 1, then what is the value of log(a+b)(a2 - b2)?​

Answer 1

Given data:

• $log\{(a+b)(a-b)\}=1$

To find:

• The value of $log\{(a+b)(a^{2}-b^{2})\}$

Step-by-step explanation:

Now, $log\{(a+b)(a-b)\}=1$

$\Rightarrow log(a^{2}-b^{2})=1$ ---(1)

Again, $log\{(a+b)(a-b)\}=1$

$\Rightarrow log(a+b)+log(a-b)=1$

$\Rightarrow log(a+b)=1-log(a-b)$ ---(2)

$\therefore log\{(a+b)(a^{2}-b^{2})\}$

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

$=1-log(a-b)+1$, by (1) and (2)

$=2-log(a-b)$

Answer:

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

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