Logarithms

Logarithms are taught at an intermediate student of algebra who is familiar with the elementary concepts.Logarithm or  Log helps us to solve many complex arithmetical problems in simpler way. Natural logarithm or ln means logarithm to the base e. e is a irrational constant equal to 2.718281828. The natural log is denoted as ln(x) or log e (x). We know that e0=1. Here in this topic we are going to see about log in algebra 2. Natural logarithm is defined for all positive real numbers.The common logarithm are to the base 10 and is written as log(x) simply without mentioning the base. If the logarithm is to any other base then it should be clearly specified.

logarithms algebra 2 - Example problems:



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logarithms algebra 2 - Problem 1:

Solve the given logarithmic expression: ln 3x + ln 5 = 3

Solution:

Given logarithmic expression: ln 3x + ln 5 = 3

ln 3x + 1.6094 = 3                   ( The value of ln 5 = 1.6094)

Subtract by 1.6094 on both side in the above expression.

ln 3x + 1.6094 - 1.6094 = 3 - 1.6094

ln 3x =  1.3906

loge 3x = 1.3906                        (ln x = loge x)

( we know, y = logbP       and P = by)

Here, y = 1.3906         P = 3x                b = e

So,    3x = e^1.3906   = 4.0172

3x = 4.0172

Divided by 3 on both side in the above equation.

` (3x)/3` = `4.0172 / 3`

x = 1.3390

Answer: The value of x = 1.3390



logarithms algebra 2 - Problem 2:

Solve the given logarithmic expression: ln 5x + ln 3x = 0

Solution:

Given logarithmic expression: ln 5x + ln 3x = 0

ln 5x + ln 3x = 0                                         ln A + ln B = ln (A + B)

ln (5x * 3x) = 0                                         (The value of ln 5 = 1.6094)

ln 15x^2 =  0

loge 15x^2 =  0                                (ln x = loge x)

( we know, y = logbP       and P = by)

Here, y = 0       P =15x^2                b = e

So,    15x^2  = e0   = 1

15x^2  = 1

Divided by 15 on both side in  the above equation.

` (15x^2)/15` = `1 / 15`

x^2 = 0.066

x = `+- sqrt 0.066`

Answer: The value of x = ± 0.2569

logarithms algebra 2 - Problem 3:
 
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Solve the given equation and find the x value. 3^-x^2 = `1/27.`

Solution:

Given equation 3^-x^2 = `1/27` .                                  exponential form

log3`(1/27)` = -x^2                                                  logarithmic form

-x^2 = log3`(1/27)`

= log3 1 -  log3 27

= 0 - 3

= -3

x = 31/2.

Answer: The value of x = 31/2.

logarithms algebra 2 - Problem 4:

Solve the given equation and find the x value:  4 (a^-n) = 160

Solution:

Given equation,   4 (a^-n) = 160

a^-n = 40                                                    exponential form

log1.140 = 2x+3                                                logarithmic form

2x + 3 = `(ln 40)/(ln 1.1) `

= 38.7039 (approximately)

Subtract by 3 on both side in above equation

2x + 3 - 3 = 38.7039 - 3

2x = 35.7039

Both side divided by 2. So, we get

`(2x)/2` =` 35.7039/2`

x = 17.85

Answer: The value of x = 17.85 (approximately)