Question

Solve the equation
${({a^4} - 2{a^2}{b^2} + {b^4})^{x - 1}} = {(a - b)^{2x}}{(a + b)^{ - 2}}$

Answer 1

Hint: Here we will apply the properties of logarithms and simplify the equation to find the value of x.

Complete step-by-step answer:
We have to evaluate ${({a^4} - 2{a^2}{b^2} + {b^4})^{x - 1}} = {(a - b)^{2x}}{(a + b)^{ - 2}}$
So taking log on both sides we will have
$\Rightarrow$ $\log {({a^4} - 2{a^2}{b^2} + {b^4})^{x - 1}} = \log ({(a - b)^{2x}}{(a + b)^{ - 2}})$
Now using the property of log that ${\text{log }}{{\text{a}}^b} = b{\text{ loga}}$ and $\log (ab) = \log a + \log b$
We will get,
$\Rightarrow$ $(x - 1)\log ({a^4} - 2{a^2}{b^2} + {b^4}) = 2x\log (a - b) - 2\log (a + b)$
Now $({a^4} - 2{a^2}{b^2} + {b^4}) = {({a^2} - {b^2})^2}$ so using it, we will get
$\Rightarrow$ $(x - 1)\log {({a^2} - {b^2})^2} = 2x\log (a - b) - 2\log (a + b)$
Again using the above mentioned log property
$\Rightarrow$ $2(x - 1)\log ({a^2} - {b^2}) = 2x\log (a - b) - 2\log (a + b)$
Now, ${a^2} - {b^2} = (a - b)(a + b)$
Using it, we get
$\Rightarrow$ $2(x - 1)\log ((a - b)(a + b)) = 2x\log (a - b) - 2\log (a + b)$
Using multiplication log property mentioned above, we get
$\Rightarrow 2(x - 1)(\log (a - b) + \log (a + b)) = 2x\log (a - b) - 2\log (a + b)$
On simplifying, we get
$\Rightarrow$ $2x(\log (a + b)) + 2x\log (a - b) - 2\log (a + b) - 2\log (a - b) = 2x\log (a - b) - 2\log (a + b)$
On simplifying, we get
$\Rightarrow$ $x\log (a + b) = \log (a - b)$
Hence, $x = \dfrac{{\log (a - b)}}{{\log (a + b)}}$

Note: While solving such equations it is always advisable to take log both sides, then having the knowledge of properties of log can help simplify the problem further and eventually reaching up to the answer.