Example

3(r+2s)=2t-4
a⋅(b-2)=3b
3(r+2s)=2t-4
a⋅(b-2)=3b

Question

$$(x^{a-b})^{a+b}(x^{b-c})^{b+c}(x^{c-a})^{c+a}=1$$

Answer

$$x=1^(1/((a-b)*(a+b)+(b-c)*(b+c)+(c-a)*(c+a)))$$

Solution


Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[{x}^{(a-b)(a+b)}{({x}^{b-c})}^{b+c}{({x}^{c-a})}^{c+a}=1\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[{x}^{(a-b)(a+b)}{x}^{(b-c)(b+c)}{({x}^{c-a})}^{c+a}=1\]
Use Power Rule: \({({x}^{a})}^{b}={x}^{ab}\).
\[{x}^{(a-b)(a+b)}{x}^{(b-c)(b+c)}{x}^{(c-a)(c+a)}=1\]
Use Product Rule: \({x}^{a}{x}^{b}={x}^{a+b}\).
\[{x}^{(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)}=1\]
Take the \(((a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a))\)th root of both sides.
\[x=\sqrt[(a-b)(a+b)+(b-c)(b+c)+(c-a)(c+a)]{1}\]
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