5.3 Properties of Logarithms

Use the change of base formula to rewrite and evaluate logs

Use properties of logs to evaluate or rewrite log expressionsUse properties of logarithms to expand or condense

logarithmic expressionsUse logarithmic functions to model and solve real-life

problems.

Change of Base Formula

Base b Base e

logax= logbx logax= ln x

logba ln a

Using the Change of Base FormulaExamples—

log4 25 =

Rewrite as:log 25log 4

1.39794.060206

= 2.3219

log2 12=

Rewrite as:log 12log 2

1.07918.30103

=3.5850

The same 2 problems can be done using ln.

Properties of Logarithms

• Product Property: loga (uv) = loga u + loga v

• Quotient Property: loga (u/v) = loga u - loga v

• Power Property: loga un = n loga u

Using Properties of Logs to find the exact value of the expression

Examplelog5 35

ln e6 – ln e2

Rewrite--log5 (5)1/3

Bring exponent out front.1/3 log5 (5)

= 1/3

Bring exponents out front.

6ln e – 2ln e

So--

6 – 2 = 4

OR we could have rewritten this as division—

Ln e6 = lne4 = 4lne = 4 e2

Using Properties of Logarithms to expand the expression as a sum,

difference and/or constantln 2/27 = ln 2 - ln 27

log310z = log310 + log3z

ln 6

x2 + 1

log 4x2y = log 4 + log x2 + log y = log 4 + 2log x + log y

= ln 6 – ln (x2 + 1)1/2

= ln 6 – 1/2ln (x2 + 1)

Write the expression as a single logarithm (Go Backwards)

ln y + ln t

= ln yt log 8 – log t = log 8/t

-4ln 2xt = ln (2xt)-4

2 ln 8 + 5 ln (x – 4) = ln 82 + ln (x – 4)5

= ln 82(x – 4)5

1/3[log x + log (x + 1)] =[log x(x + 1)]1/3

2[3ln x – ln (x + 1) – ln(x – 1)] =[3ln x – ln (x + 1) – ln(x – 1)]2

= ln x3 2

(x + 1)(x – 1) Foil this