Logarithmic function

Logarithmic function

Done By:Al-HanoofAmnaDana Ghada

Logarithmic Function

Changing from Exponential to Logarithmic form. ( Ghada )

Graphing of Logarithmic Function.( Amna ) Common Logarithm.( Ghada ) Natural Logarithm.( Ghada ) Laws of Logarithmic Function. ( Dana ) Change of the base.( Ghada ) Solving of Logarithmic Function.( Al-Hanoof ) Application of Logarithmic Function.(Al-Hanoof)


Every exponential function f(x) = a x, with a > 0 and a ≠ 1. is a one-to-one function, therefore has an inverse function(f-1). The inverse function is called the Logarithmic function with base a and is denoted by Loga

Let a be a positive number with a ≠ 1. The logarithmic function with base a, denoted by loga is defined by:

Loga x = y a y = Х Clearly, Loga Х is the exponent to which the base a must be raised to give Х


Logarithmic form Exponential form

Exponent Exponent

Loga x = y a^ y = Х

Base Base


Example:

Logarithmic form

Log2 8= 3

Exponential form

2^3=8


Graphs of Logarithmic Functions:

The exponential function f(x) =a^x has

Domain: IR

Range: (0.∞),

Since the logarithmic function is the inverse function for the exponential function , it has

Domain : (0, ∞)

Range: IR.


The graph of f(x) = Loga x is obtained by reflecting the graph of f(x) = a^ x the line y = x

x-intercept of the function y = Loga x is 1

f(x) = a^ x

y = x


This is the basic function y= Loga x

y = loga x

1 2 3 4-1-2-3-4

x1

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y =- loga x

The function is reflected in the x-axis .

1 2 3 4-1-2-3-4

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y = log2 (-x)

The function is reflected in the y-axis .

1 2 3 4-1-2-3-4

x1

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The function is shifted to the left by two unites .

Y=loga(x+2)

1 2 3 4-1-2-3-4

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The function is shifted to the right by two unites .

y = loga (x-2)

1 2 3 4-1-2-3-4

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The function is shifted to the upward by two unites .

y = logax +2

1 2 3 4-1-2-3-4

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y = loga x -2

The function is shifted to the downward by two unites .

1 2 3 4-1-2-3-4

x1

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Example:Finding the domain of a logarithmic function: F(x)=log(x-2)Solution:As any logarithmic function lnx is defined when x>0,

thus, the domain of f(x) is x-2 >0 X>2 So the domain =(2,∞)


Common Logarithmic;

The logarithm with base 10 is called the common logarithm and is denoted by omitting the base:

log x = log10x

Natural Logarithms:

The logarithm with base e is called the natural logarithm and is denoted by In:

ln x =logex


The natural logarithmic function y = In x is the inverse function of the exponential function y = e^X.By the definition of inverse functions we have:

ln x =y e^y=x

1 2 3 4-1-2-3-4

x1

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Y=e^x

Y=ln x

Laws of logarithms:

Let a be a positive number, with a≠1. let A>0, B>0, and C be any real numbers.

1. loga (AB) = loga A + loga B

log2 (6x) = log2 6 + log2 x

2. loga (A/B) = loga A - loga B

log2 (10/3) = log2 10 – log2 3

3. loga A^c = C loga A

log3 √5 = log3 51/2 = 1/2 log3 5


Rewrite each expression using logarithm laws

log5 (x^3 y^6)

= log5 x^3 + log5 y^6 law1

= 3 log5 x + 6 log5 y law3

ln (ab/3√c)

= ln (ab) – ln 3√c law2

= ln a + ln b – ln c1/3 law1

= ln a + ln b – 1/3 ln c law3


Express as a single logarithm

3 log x + ½ log (x+1)

= log x^3 + log (x+1)^1/2 law3

=log x^3(x+1)^1/2 law1

3 ln s + ½ ln t – 4 ln (t2+1)

= ln s^3 + ln t^1/2 – ln (t^2+1)^4 law3

= ln ( s^3 t^1/2) – ln (t^2 + 1)^4 law1

= ln s^3 √t /(t2+1)^4 law2



*WARNING:

loga (x+y) ≠ logax +logay

Log 6/log2 ≠ log(6/2)

(log2x)3 ≠ 3log2x


Change of Base:Sometimes we need to change from logarithms in one

base to logarithms in another base. b^y = x (exponential form)

logab^y = logax (take loga for both sides)

y log a b =logax (law3)

y=(loga x)/(loga b) (divide by logab)


Example:

Since all calculators are operational for log10 we will change the base to 10

Log8 5 = log10 5/ log10 8≈ 0.77398 (approximating the answer by using the calculator)


Solving the logarithmic Equations:Example:Find the solution of the equation log 3^(x+2) = log7.SOLUTION:

) x + 2 (log 3=log7 (bring down the exponent)X+2= log7 (divide by log 3 ) log 3x = log7 -2 (subtract by 2) log3


Application of e and Exponential Functions:In the calculation of interest exponential function is used. In order to

make the solution easier we use the logarithmic function.A= P (1+ r/n)^nt

A is the money accumulated.P is the principal (beginning) amount r is the annual interest rate n is the number of compounding periods per yeart is the number of years

There are three formulas:

A = p(1+r) Simple interest (for one year)

A(t) = p(1+r/n)nt Interest compounded n times per year

A(t) = pert Interest compounded continuously


Example: A sum of $500 is invested at an interest rate 9%per year. Find the time required for the money to double if the interest is compounded according to the following method.

a) Semiannual b) continuous Solution:

)a( We use the formula for compound interest with P = $5000, A (t) = $10,000r = 0.09, n = 2, and solve the resulting exponential equation for t.

(1.045)^2t = 2 (Divide by 5000)log (1.04521)^2t = log 2 (Take log of each side)2t log 1.045 = log 2 Law 3 (bring down the exponent)t= (log 2)/ (2 log 1.045) (Divide by 2 log 1.045)t ≈ 7.9 The money will double in 7.9 years. (using a calculator)


(b) We use the formula for continuously compounded interest with P = $5000,

A(t) = $10,000, r = 0.09, and solve the resulting exponential equation

for t.

5000e^0.09t = 10,000

e^0.091 = 2 (Divide by 5000)

In e0.091 = In 2 (Take 10 of each side)

0.09t = In 2 (Property of In)

t=(In 2)/(0.09) (Divide by 0.09)

t ≈7.702 (Use a calculator)

The money will double in 7.7 years.

Date post:	17-Jan-2016
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Logarithmic function

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