To solve the given logarithmic equation, we will rewrite it in exponential form using the definition of a logarithm.

lo g_{b} x = y \Leftrightarrow x = b^{y}

This definition tells us how to rewrite the logarithm equivalent of

y

in exponential form. The argument

x

is equal to

b

raised to the power of

y .

lo g_{2} (x^{2} - x - 6) = 2 ⇕ x^{2} - x - 6 = 2^{2}

We can see above that

2

is the exponent to which

2

must be raised to obtain

x^{2} - x - 6 .

Now, let's simplify our equation.

x^{2} - x - 6 = 2^{2}

CalcPow

Calculate power

x^{2} - x - 6 = 4

SubEqn

$LHS - 4 = RHS - 4$

x^{2} - x - 10 = 0

We will use the Quadratic Formula to solve the given quadratic equation.

a x^{2} + b x + c = 0 \Leftrightarrow x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

We first need to identify the values of

a,

b,

and

c .

x^{2} - x - 10 = 0 ⇕ 1 x^{2} + (- 1) x + (- 10) = 0

We see that

a = 1,

b = - 1,

and

c = - 10 .

Let's substitute these values into the Quadratic Formula.

x = \frac{- b \pm b ^{2} - 4 a c}{2 a}

SubstituteValues

Substitute values

x = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( 1 ) ( - 1 0 )}{2 ( 1 )}

▼

Solve for

x

and Simplify

NegNeg

$- (- a) = a$

x = \frac{1 \pm ( - 1 ) ^{2} - 4 ( 1 ) ( - 1 0 )}{2 ( 1 )}

CalcPow

Calculate power

x = \frac{1 \pm 1 - 4 ( 1 ) ( - 1 0 )}{2 ( 1 )}

MultByOne

$a \cdot 1 = a$

x = \frac{1 \pm 1 - 4 ( - 1 0 )}{2}

MultNegNegOnePar

$- a (- b) = a \cdot b$

x = \frac{1 \pm 1 + 4 0}{2}

AddTerms

Add terms

x = \frac{1 \pm 4 1}{2}

Using the Quadratic Formula, we found that the solutions of the given equation are

x = \frac{1 \pm 4 1}{2} .

Therefore, the exact solutions are

x_{1} = \frac{1 + 4 1}{2}

and

x_{2} = \frac{1 - 4 1}{2} .

Let's use a calculator to approximate these values.

x_{1} x_{2} = \frac{1 + 4 1}{2} \approx 3.702 = \frac{1 - 4 1}{2} \approx - 2.702

To check our answer, we will substitute both

3.702

and

- 2.702

for

x

in the given equation one at a time.

lo g_{2} (x^{2} - x - 6) = 2

$x \approx 3.702$

lo g_{2} (3.702^{2} - 3.702 - 6) \approx ? 2

CalcPow

Calculate power

lo g_{2} (13.704804 - 3.702 - 6) \approx ? 2

SubTerms

Subtract terms

lo g_{2} 4.002804 \approx ? 2

Calculate logarithm

2.001010975 \approx 2 ✓

Since substituting

3.702

for

x

in the given equation produces a true statement,

x \approx 3.702

is a solution to our equation. Let's now check for the

x \approx - 2.702 .

lo g_{2} (x^{2} - x - 6) = 2

$x \approx - 2.702$

lo g_{2} ((- 2.702)^{2} - (- 2.702) - 6) \approx ? 2

CalcPow

Calculate power

lo g_{2} (7.300804 - (- 2.702) - 6) \approx ? 2

NegNeg

$- (- a) = a$

lo g_{2} (7.300804 + 2.702 - 6) \approx ? 2

AddSubTerms

Add and subtract terms

lo g_{2} 4.002804 \approx ? 2

Calculate logarithm

2.001010975 \approx 2 ✓

Substituting

- 2.702

for

x

in the given equation produces a true statement. Therefore,

x \approx - 2.702

is also a solution to our equation.

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