Exercise 26 Page 327

a We are asked to evaluate the following expression.

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

Let's rewrite the logarithm's arguments as powers, and then use the properties of logarithms to further simplify.

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

Write as a power

\frac{lo g _{2} 2 ^{5}}{lo g _{2} 2 ^{2}}

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

\frac{5 lo g _{2} 2}{2 lo g _{2} 2}

$\frac{a}{b} = \frac{a / lo g _{2} 2}{b / lo g _{2} 2}$

\frac{5}{2}

b We are asked to solve the following logarithm.

\frac{lo g 3 2}{lo g 4}

Similar to Part A, we will rewrite the logarithms by changing the argument and then simplify using the properties of logarithms.

\frac{lo g 3 2}{lo g 4}

Write as a power

\frac{lo g 2 ^{5}}{lo g 2 ^{2}}

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

\frac{5 lo g 2}{2 lo g 2}

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

\frac{5}{2}

c Notice that the answers from Part A and Part B are the same. To explain why, we should have a look at the change of base formula.

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

What happens if we apply this formula on both of our expressions?

\frac{lo g _{2} 3 2}{lo g _{2} 4} \frac{lo g 3 2}{lo g 4} = lo g_{4} 32 = lo g_{4} 32

Since the right-hand sides are equivalent, the fractions from Parts A and B give the same results.

d We are asked to change

lo g_{2} 7

into a logarithmic expression using a base of

5 .

Let's label the result of the logarithmic expression

x .

We can then write the following equation.

lo g_{2} 7 = x

Next, using the definition of a logarithm we will rewrite the logarithmic expression as a power.

lo g_{2} 7 = x ⇕ 2^{x} = 7

Now, by setting the log base of

5

to both sides, we can rewrite it as a logarithmic expression using a base of

5 .

2^{x} = 7

$lo g_{5} (LHS) = lo g_{5} (RHS)$

lo g_{5} 2^{x} = lo g_{5} 7

$lo g_{5} (a^{m}) = m \cdot lo g_{5} (a)$

x lo g_{5} 2 = lo g_{5} 7

$LHS / lo g_{5} 2 = RHS / lo g_{5} 2$

x = \frac{lo g _{5} 7}{lo g _{5} 2}

Since

x

equals both

\frac{l o g _{5} 7}{l o g _{5} 2}

and

lo g_{2} 7,

we can equate these expressions.

lo g_{2} 7 = \frac{lo g _{5} 7}{lo g _{5} 2}

7.1.1 Using Logarithms to Solve Exponential Equations p.323

7.1.2 Investigating the Properties of Logarithms p.327-328

7.1.3 Writing Equations of Exponential Functions p.331-332

7.1.4 An Application of Logarithms p.336-337