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_{2} x + lo g_{2} (x - 2) = 3

$lo g_{2} (m) + lo g_{2} (n) = lo g_{2} (m n)$

lo g_{2} (x (x - 2)) = 3

$m = lo g_{2} (2^{m})$

lo g_{2} (x (x - 2)) = lo g_{2} 2^{3}

CalcPow

Calculate power

lo g_{2} (x (x - 2)) = lo g_{2} 8

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 solve for

x

by factoring.

lo g_{2} (x (x - 2)) = lo g_{2} 8

Equate arguments

x (x - 2) = 8

▼

Solve for

x

Distr

Distribute $x$

x^{2} - 2 x = 8

SubEqn

$LHS - 8 = RHS - 8$

x^{2} - 2 x - 8 = 0

Rewrite

Rewrite $- 2 x$ as $2 x - 4 x$

x^{2} + 2 x - 4 x - 8 = 0

FactorOut

Factor out $x$

x (x + 2) - 4 x - 8 = 0

FactorOut

Factor out $- 4$

x (x + 2) - 4 (x + 2) = 0

FactorOut

Factor out $(x + 2)$

(x + 2) (x - 4) = 0

ZeroProdProp

Use the Zero Product Property

x + 2 = 0 x - 4 = 0 (I) (II)

SubEqn

$(I):$ $LHS - 2 = RHS - 2$

x = - 2 x - 4 = 0

AddEqn

$(II):$ $LHS + 4 = RHS + 4$

x = - 2 x = 4

To check for extraneous solutions, we will substitute both

- 2

and

4

for

x

in the given equation one at a time.

lo g_{2} x + lo g_{2} (x - 2) = 3

Substitute

$x = - 2$

lo g_{2} (- 2) + lo g_{2} (- 2 - 2) = ? 3

SubTerm

Subtract term

lo g_{2} (- 2) + lo g_{2} (- 4) = ? 3 \times

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

lo g_{2} (- 2)

and

lo g_{2} (- 4)

are undefined, and

- 4

is an extraneous solution. Let's now check for

x = 4 .

lo g_{2} x + lo g_{2} (x - 2) = 3

Substitute

$x = 4$

lo g_{2} 4 + lo g_{2} (4 - 2) = ? 3

SubTerm

Subtract term

lo g_{2} 4 + lo g_{2} 2 = ? 3

$lo g_{2} (2) = 1$

lo g_{2} 4 + 1 = ? 3

Calculate logarithm

2 + 1 = ? 3

AddTerms

Add terms

3 = 3 ✓

Since substituting

4

for

x

in the given equation produces a true statement,

x = 4

is the solution to our equation.

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