Solve for x natural log of x^2-x+17=3/4

$\ln (x^{2} - x + 17) = \frac{3}{4}$

Step 1

To solve for $x$ , rewrite the equation using properties of logarithms.

$e^{\ln (x^{2} - x + 17)} = e^{\frac{3}{4}}$

Step 2

Rewrite $\ln (x^{2} - x + 17) = \frac{3}{4}$ in exponential form using the definition of a logarithm. If $x$ and $b$ are positive real numbers and $b \neq 1$ , then $\log_{b} (x) = y$ is equivalent to $b^{y} = x$ .

$e^{\frac{3}{4}} = x^{2} - x + 17$

Step 3

Solve for $x$ .

Rewrite the equation as $x^{2} - x + 17 = e^{\frac{3}{4}}$ .

$x^{2} - x + 17 = e^{\frac{3}{4}}$

Subtract $e^{\frac{3}{4}}$ from both sides of the equation.

$x^{2} - x + 17 - e^{\frac{3}{4}} = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = - 1$ , and $c = 17 - e^{\frac{3}{4}}$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (1 \cdot (17 - e^{\frac{3}{4}}))}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 1 \cdot (17 - e^{\frac{3}{4}})}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot (17 - e^{\frac{3}{4}})}}{2 \cdot 1}$

Apply the distributive property.

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 17 - 4 (- e^{\frac{3}{4}})}}{2 \cdot 1}$

Multiply $- 4$ by $17$ .

$x = \frac{1 \pm \sqrt{1 - 68 - 4 (- e^{\frac{3}{4}})}}{2 \cdot 1}$

Multiply $- 1$ by $- 4$ .

$x = \frac{1 \pm \sqrt{1 - 68 + 4 e^{\frac{3}{4}}}}{2 \cdot 1}$

Subtract $68$ from $1$ .

$x = \frac{1 \pm \sqrt{- 67 + 4 e^{\frac{3}{4}}}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{1 \pm \sqrt{- 67 + 4 e^{\frac{3}{4}}}}{2}$

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{- 67 + 4 e^{\frac{3}{4}}}}{2}, \frac{1 - \sqrt{- 67 + 4 e^{\frac{3}{4}}}}{2}$

Step 4

Exclude the solutions that do not make $\ln (x^{2} - x + 17) = \frac{3}{4}$ true.

$No solution$

Solve for x natural log of x^2-x+17=3/4

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