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_{3} (x - 9) + lo g_{3} (x - 3) = 2

$lo g_{6} (m) + lo g_{6} (n) = lo g_{6} (m n)$

lo g_{3} ((x - 9) (x - 3)) = 2

Distr

Distribute $(x - 3)$

lo g_{3} (x (x - 3) - 9 (x - 3)) = 2

▼

Distribute

x & - 9

Distr

Distribute $x$

lo g_{3} ((x^{2} - 3 x) - 9 (x - 3)) = 2

Distr

Distribute $- 9$

lo g_{3} (x^{2} - 3 x - 9 x + 27) = 2

SubTerm

Subtract term

lo g_{3} (x^{2} - 12 x + 27) = 2

$m = lo g_{3} (3^{m})$

lo g_{3} (x^{2} - 12 x + 27) = lo g_{3} 3^{2}

CalcPow

Calculate power

lo g_{3} (x^{2} - 12 x + 27) = lo g_{3} 9

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_{3} (x^{2} - 12 x + 27) = lo g_{3} 9

Equate arguments

x^{2} - 12 x + 27 = 9

SubEqn

$LHS - 9 = RHS - 9$

x^{2} - 12 x + 18 = 0

We will use the Quadratic Formula to solve the obatained 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} - 12 x + 18 = 0 ⇕ 1 x^{2} + (- 12) x + 18 = 0

We see that

a = 1,

b = - 12,

and

c = 18 .

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 2 ) \pm ( - 1 2 ) ^{2} - 4 ( 1 ) ( 1 8 )}{2 ( 1 )}

▼

Solve for

x

and Simplify

NegNeg

$- (- a) = a$

x = \frac{1 2 \pm ( - 1 2 ) ^{2} - 4 ( 1 ) ( 1 8 )}{2 ( 1 )}

CalcPow

Calculate power

x = \frac{1 2 \pm 1 4 4 - 4 ( 1 ) ( 1 8 )}{2 ( 1 )}

MultByOne

$a \cdot 1 = a$

x = \frac{1 2 \pm 1 4 4 - 4 ( 1 8 )}{2}

Multiply

x = \frac{1 2 \pm 1 4 4 - 7 2}{2}

SubTerm

Subtract term

x = \frac{1 2 \pm 7 2}{2}

SplitIntoFactors

Split into factors

x = \frac{1 2 \pm 3 6 \cdot 2}{2}

SqrtProd

$a \cdot b = a \cdot b$

x = \frac{1 2 \pm 3 6 \cdot 2}{2}

CalcRoot

Calculate root

x = \frac{1 2 \pm 6 \cdot 2}{2}

FactorOut

Factor out $6$

x = \frac{6 ( 2 \pm 2 )}{2}

ReduceFrac

$\frac{a}{b} = \frac{a / 2}{b / 2}$

x = 3 (2 \pm 2)

Distr

Distribute $3$

x = 6 \pm 32

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

x = 6 \pm 32 .

Therefore, the exact solutions are

x_{1} = 6 + 32

and

x_{2} = 6 - 32 .

Let's use a calculator to approximate these values.

x_{1} x_{2} = 6 + 32 \approx 10.24 = 6 - 32 \approx 1.76

To check for extraneous solutions, we will substitute both

10.24

and

1.76

for

x

in the given equation one at a time.

lo g_{3} (x - 9) + lo g_{3} (x - 3) = 2

Substitute

$x = 10.24$

lo g_{3} (10.24 - 9) + lo g_{3} (10.24 - 3) = ? 2

SubTerms

Subtract terms

lo g_{3} 1.24 + lo g_{3} 7.24 = ? 2

Calculate logarithm

0.2 + 1.8 = ? 2

AddTerms

Add terms

2 = 2 ✓

Since substituting

10.24

for

x

in the given equation produces a true statement,

x = 10.24

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

x = 1.76 .

lo g_{3} (x - 9) + lo g_{3} (x - 3) = 2

Substitute

$x = 1.76$

lo g_{3} (1.76 - 9) + lo g_{3} (1.76 - 3) = ? 2

SubTerms

Subtract terms

lo g_{3} (- 7.24) + lo g_{3} (- 1.24) = ? 2 \times

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

lo g_{3} (- 7.24)

and

lo g_{3} (- 1.24)

are undefined, and

1.76

is an extraneous solution.

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