Write in Exponential Form log base 2x of x^2+5x-16=7/5

$\log_{2 x} (x^{2} + 5 x - 16) = \frac{7}{5}$

Step 1

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, $b = 2 x$ , $x = x^{2} + 5 x - 16$ , and $y = \frac{7}{5}$ .

$b = 2 x$

$x = x^{2} + 5 x - 16$

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

Step 2

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

$2 x^{\frac{7}{5}} = x^{2} + 5 x - 16$

Write in Exponential Form log base 2x of x^2+5x-16=7/5

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