We want to solve an equation involving more than one logarithm. To do so, we will use the Product Property of Logarithms.

lo g_{b} m n = lo g_{b} m + lo g_{b} n

First, we will isolate the logarithm that contains the variable. Then, we can use the above property to isolate the variable from the logarithm.

lo g_{5} (x + 4) + lo g_{5} (x + 1) = 2

$lo g_{5} (m) + lo g_{5} (n) = lo g_{5} (m n)$

lo g_{5} ((x + 4) (x + 1)) = 2

Distr

Distribute $(x + 1)$

lo g_{5} (x (x + 1) + 4 (x + 1)) = 2

▼

Distribute

x & 4

Distr

Distribute $x$

lo g_{5} ((x^{2} + x) + 4 (x + 1)) = 2

Distr

Distribute $4$

lo g_{5} (x^{2} + x + 4 x + 4) = 2

AddTerms

Add terms

lo g_{5} (x^{2} + 5 x + 4) = 2

$m = lo g_{5} (5^{m})$

lo g_{5} (x^{2} + 5 x + 4) = lo g_{5} 5^{2}

CalcPow

Calculate power

lo g_{5} (x^{2} + 5 x + 4) = lo g_{5} 25

Next, we will use the fact that if two equivalent logarithmic expressions have the same base, then the arguments must be equal.

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

Let's apply the above property to our equation and then simplify it.

lo g_{5} (x^{2} + 5 x + 4) = lo g_{5} 25

Equate arguments

x^{2} + 5 x + 4 = 25

SubEqn

$LHS - 25 = RHS - 25$

x^{2} + 5 x - 21 = 0

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

a x^{2} + b x + c = 0 ⇕ 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} + 5 x - 21 = 0 ⇕ 1 x^{2} + 5 x + (- 21) = 0

We see that

a = 1,

b = 5,

and

c = - 21 .

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{- 5 \pm 5 ^{2} - 4 ( 1 ) ( - 2 1 )}{2 ( 1 )}

▼

Solve for

x

and Simplify

CalcPow

Calculate power

x = \frac{- 5 \pm 2 5 - 4 ( 1 ) ( - 2 1 )}{2 ( 1 )}

MultByOne

$a \cdot 1 = a$

x = \frac{- 5 \pm 2 5 - 4 ( - 2 1 )}{2}

MultNegNegOnePar

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

x = \frac{- 5 \pm 2 5 + 8 4}{2}

AddTerms

Add terms

x = \frac{- 5 \pm 1 0 9}{2}

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

x = \frac{- 5 \pm 1 0 9}{2} .

Therefore, the exact solutions are

x_{1} = \frac{- 5 + 1 0 9}{2}

and

x_{2} = \frac{- 5 - 1 0 9}{2} .

Let's use a calculator to approximate these values.

x_{1} x_{2} = \frac{- 5 + 1 0 9}{2} \approx 2.72 = \frac{- 5 - 1 0 9}{2} \approx - 7.72

To check for extraneous solutions, we will substitute both

2.72

and

- 7.72

for

x

in the given equation one at a time.

lo g_{5} (x + 4) + lo g_{5} (x + 1) = 2

Substitute

$x = 2.72$

lo g_{5} (2.72 + 4) + lo g_{5} (2.72 + 1) = ? 2

AddTerms

Add terms

lo g_{5} 6.72 + lo g_{5} 3.72 = ? 2

Calculate logarithm

1.18 + 0.82 = ? 2

AddTerms

Add terms

2 = 2 ✓

Since substituting

2.72

for

x

in the given equation produces a true statement,

x = 2.72

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

x = - 7.72 .

lo g_{5} (x + 4) + lo g_{5} (x + 1) = 2

Substitute

$x = - 7.72$

lo g_{5} (- 7.72 + 4) + lo g_{5} (- 7.72 + 1) = ? 2

AddTerms

Add terms

lo g_{5} (- 3.72) + lo g_{5} (- 6.72) = ? 2 \times

The argument of a logarithmic function cannot be negative, so both

lo g_{5} (- 3.72)

and

lo g_{5} (- 6.72)

are undefined, and

- 7.72

is an extraneous solution.

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