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 equation.

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

In the above formula,

m,

b,

and

c

are positive numbers, with

b \neq = 1

and

c \neq = 1 .

We will apply this formula with

c = 2

to simplify the expressions.

lo g_{4} x = lo g_{8} 4 x

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

\frac{lo g _{2} x}{lo g _{2} 4} = \frac{lo g _{2} 4 x}{lo g _{2} 8}

Calculate logarithm

\frac{lo g _{2} x}{2} = \frac{lo g _{2} 4 x}{3}

Cross multiply

3 lo g_{2} x = 2 lo g_{2} 4 x

$m \cdot lo g_{2} (a) = lo g_{2} (a^{m})$

lo g_{2} x^{3} = lo g_{2} (4 x)^{2}

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_{2} x^{3} = lo g_{2} (4 x)^{2} \Leftrightarrow x^{3} = (4 x)^{2}

Let's now solve the equation

x^{3} = (4 x)^{2} .

x^{3} = (4 x)^{2}

$(a \cdot b)^{m} = a^{m} \cdot b^{m}$

x^{3} = 16 x^{2}

$LHS - 16 x^{2} = RHS - 16 x^{2}$

x^{3} - 16 x^{2} = 0

Factor out $x^{2}$

x^{2} (x - 16) = 0

Use the Zero Product Property

x^{2} = 0 x - 16 = 0 (I) (II)

$(I):$ $LHS = RHS$

x = 0 x - 16 = 0

$(II):$ $LHS + 16 = RHS + 16$

x = 0 x = 16

To check our answer, we will substitute both

0

and

16

for

x

in the given equation one at a time.

lo g_{4} x = lo g_{8} 4 x

$x = 0$

lo g_{4} 0 = ? lo g_{8} 4 (0)

Zero Property of Multiplication

lo g_{4} 0 = ? lo g_{8} 0 \times

Note that you can never get zero by raising a number different from zero to any power, so both

lo g_{4} 0,

and

lo g_{8} 0

are undefined. Therefore,

0

is an extraneous solution. Let's now check for

x = 16 .

lo g_{4} x = lo g_{8} 4 x

$x = 16$

lo g_{4} 16 = ? lo g_{8} 4 (16)

lo g_{4} 16 = ? lo g_{8} 64

Calculate logarithm

2 = 2 ✓

Since substituting

16

for

x

in the given equation produces a true statement,

x = 16

is the solution to our equation.

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