Use this formula: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[{x}^{4}-38{x}^{r}+1=0thenWhat\imath sthevalueof\times \frac{({x}^{4}+1)({x}^{4}-1)}{{x}^{4}}\]
Rewrite \({x}^{4}-1\) in the form \({a}^{2}-{b}^{2}\), where \(a={x}^{2}\) and \(b=1\).
\[{x}^{4}-38{x}^{r}+1=0thenWhat\imath sthevalueof\times \frac{({x}^{4}+1)({({x}^{2})}^{2}-{1}^{2})}{{x}^{4}}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[{x}^{4}-38{x}^{r}+1=0thenWhat\imath sthevalueof\times \frac{({x}^{4}+1)({x}^{2}+1)({x}^{2}-1)}{{x}^{4}}\]
Rewrite \({x}^{2}-1\) in the form \({a}^{2}-{b}^{2}\), where \(a=x\) and \(b=1\).
\[{x}^{4}-38{x}^{r}+1=0thenWhat\imath sthevalueof\times \frac{({x}^{4}+1)({x}^{2}+1)({x}^{2}-{1}^{2})}{{x}^{4}}\]
Use Difference of Squares: \({a}^{2}-{b}^{2}=(a+b)(a-b)\).
\[{x}^{4}-38{x}^{r}+1=0thenWhat\imath sthevalueof\times \frac{({x}^{4}+1)({x}^{2}+1)(x+1)(x-1)}{{x}^{4}}\]
Simplify \(0thenWhat\imath sthevalueof\times \frac{({x}^{4}+1)({x}^{2}+1)(x+1)(x-1)}{{x}^{4}}\) to \(0\).
\[{x}^{4}-38{x}^{r}+1=0\]
Subtract \({x}^{4}\) from both sides.
\[-38{x}^{r}+1=-{x}^{4}\]
Subtract \(1\) from both sides.
\[-38{x}^{r}=-{x}^{4}-1\]
Divide both sides by \(-38\).
\[{x}^{r}=-\frac{-{x}^{4}-1}{38}\]
Use Definition of Common Logarithm: \({b}^{a}=x\) if and only if \(log_b(x)=a\).
\[r=\log_{x}{(-\frac{-{x}^{4}-1}{38})}\]
r=log(x,-(-x^4-1)/38)
