Write in Exponential Form log base (x-2) of x^2-x+14=7/5

$\log_{(x - 2)} (x^{2} - x + 14) = \frac{7}{5}$

For logarithmic equations, $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ such that $x > 0$ , $b > 0$ , and $b \neq 1$ . In this case, $b = x - 2$ , $x = x^{2} - x + 14$ , and $y = \frac{7}{5}$ .

$b = x - 2$

$x = x^{2} - x + 14$

$y = \frac{7}{5}$

Substitute the values of $b$ , $x$ , and $y$ into the equation $b^{y} = x$ .

${(x - 2)}^{\frac{7}{5}} = x^{2} - x + 14$

Write in Exponential Form log base (x-2) of x^2-x+14=7/5

Related Questions

Support LCMGCF.com by disabling your adblocker.