We want to solve an equation involving more than one logarithm. To do so, we will use the Change of Base Formula to evaluate the given expression.

lo g_{b} m = \frac{lo g _{c} m}{lo g _{c} b}

Numbers

m,

b,

and

c

are positive numbers, with

b \neq = 1

and

c \neq = 1 .

We will apply this formula with

c = 9

to simplify one of the expressions.

lo g_{3} (x - 6) = lo g_{9} 2 x

$lo g_{c} a = \frac{lo g _{b} a}{lo g _{b} c}$

\frac{lo g _{9} ( x - 6 )}{lo g _{9} 3} = lo g_{9} 2 x

Calculate logarithm

\frac{lo g _{9} ( x - 6 )}{1 / 2} = lo g_{9} 2 x

DivByFracD

$\frac{a}{b / c} = \frac{a \cdot c}{b}$

lo g_{9} (x - 6) \cdot 2 = lo g_{9} 2 x

CommutativePropMult

Commutative Property of Multiplication

2 lo g_{9} (x - 6) = lo g_{9} 2 x

$m \cdot lo g_{9} (a) = lo g_{9} (a^{m})$

lo g_{9} (x - 6)^{2} = lo g_{9} 2 x

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.

lo g_{9} (x - 6)^{2} = lo g_{9} 2 x

Equate arguments

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

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

x^{2} - 12 x + 36 = 2 x

SubEqn

$LHS - 2 x = RHS - 2 x$

x^{2} - 14 x + 36 = 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} - 14 x + 36 = 0 ⇕ 1 x^{2} + (- 14) x + 36 = 0

We see that

a = 1,

b = - 14,

and

c = 36 .

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 4 ) \pm ( - 1 4 ) ^{2} - 4 ( 1 ) ( 3 6 )}{2 ( 1 )}

▼

Evaluate right-hand side

NegNeg

$- (- a) = a$

x = \frac{1 4 \pm ( - 1 4 ) ^{2} - 4 ( 1 ) ( 3 6 )}{2 ( 1 )}

CalcPow

Calculate power

x = \frac{1 4 \pm 1 9 6 - 4 ( 1 ) ( 3 6 )}{2 ( 1 )}

IdPropMult

Identity Property of Multiplication

x = \frac{1 4 \pm 1 9 6 - 4 ( 3 6 )}{2}

Multiply

x = \frac{1 4 \pm 1 9 6 - 1 4 4}{2}

SubTerm

Subtract term

x = \frac{1 4 \pm 5 2}{2}

SplitIntoFactors

Split into factors

x = \frac{1 4 \pm 4 ( 1 3 )}{2}

SqrtProd

$a \cdot b = a \cdot b$

x = \frac{1 4 \pm 4 1 3}{2}

CalcRoot

Calculate root

x = \frac{1 4 \pm 2 1 3}{2}

FactorOut

Factor out $2$

x = \frac{2 ( 7 \pm 1 3 )}{2}

CancelCommonFac

Cancel out common factors

x = 7 \pm 13

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

x = 7 \pm 13 .

Therefore, the exact solutions are

x_{1} = 7 + 13

and

x_{2} = 7 - 13 .

Let's use a calculator to approximate these values.

x_{1} x_{2} = 7 + 13 \approx 10.61 = 7 - 13 \approx 3.39

To check our answer, we will substitute both

10.61

and

3.39

for

x

in the given equation one at a time.

lo g_{3} (x - 6) = lo g_{9} 2 x

Substitute

$x = 10.61$

lo g_{3} (10.61 - 6) = ? lo g_{9} 2 (10.61)

SubTerm

Subtract term

lo g_{3} 4.61 = ? lo g_{9} 2 (10.61)

Multiply

lo g_{3} 4.61 = ? lo g_{9} 21.22

Calculate logarithm

1.39 = 1.39 ✓

Since substituting

10.61

for

x

in the given equation produces a true statement,

x = 10.61

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

x = 3.39 .

lo g_{3} (x - 6) = lo g_{9} 2 x

Substitute

$x = 3.39$

lo g_{3} (3.39 - 6) = ? lo g_{9} 2 (3.39)

SubTerm

Subtract term

lo g_{3} (- 2.61) = ? lo g_{9} 2 (3.39)

Multiply

lo g_{3} (- 2.61) = ? lo g_{9} 6.78 \times

The argument of a logarithmic function cannot be negative, so

lo g_{3} (- 2.6)

is undefined, and

3.39

is an extraneous solution.

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