INTRODUCTION TO LOGARITHM (LAWS OF LOGARITHM)

Logarithms are the inverse function of exponentiation. 

This implies the logarithm of a given number $y$ is the exponent to which another fixed number, $a$ must be raised, to produce that number $y$.

Now, let's look at the three basic laws of logarithmic .

First law of logarithm

$\log _{b}s+\log_{b}t=\log_{b}st$

Second law of logarithm

$\log_{b}s-\log_{b}t=\log_{b}\frac{s}{t}$

Third law of logarithm

$\log_{b}s^t=t\log_{b}s$

Forthwith, let's try some examples.

Example 1

Solve $\log_{3}27$

Solution:

Let $\log_{3}27=x$

Index form

$27=3^x$

$3^3=3^x$

$3=x$

$x=3$

Hence

$\log_{3}27=3$

Example 2

$\log_{4}x=3$

Solution:

$\log_{4}x=3$ becomes $x=4^3$

So $x=64$

Example 3

Write $\log_{}800$ in terms of log 4, log 5, log 8 to any base.

Solution:

$\log_{}800=\log_{}(25\times32)=\log_{}(25\times 8\times4)$

$\log_{}(5^2\times8\times4)$

$\log_{}5^2+\log_{}8+\log_{}(4)$

By the first law of logarithms

$\log_{}800=2\log_{}5+\log_{}8+\log_{}4$

 By the third law of logarithm

Example 4

Simplify  $\log_{2}1024+\log_{2}256-\log_{2}512$

Solution:

$1024=2^{10}$, $256=2^8$,  $512=2^9$

Hence,

$\log_{2}1024+\log_{2}256-\log_{2}512$

$=\log_{2}2^{10}+\log_{2}2^8-\log_{2}2^9$

$=10\log_{2}2+8\log_{2}2-9\log_{2}2$

By the third law of logarithm

If $\log_{2}2=1$, Then,

$10\log_{2}2+8\log_{2}2-9\log_{2}2$

$10(1)+8(1)-9(1)$

$10+8-9=9$

Example 5

Evaluate $\log_{6}216-\log_{¼}256+\log_{5}3125$

Solution: 

The base is not the same, so we will simplify separately before evaluating

Let $\log_{6}216=a$

Index form

$216=6^a$

$6^3=6^a$

$3=a$

$a=3$

Let $\log_{¼}256=b$

Index form

$256=¼^b$

$4^4=4^{-1(b)}$

$4^4=4^{-b}$

$4=-b$

Divide both sides by $-1$

$\frac{4}{-1}=\frac{-b}{-1}$

$-4=b$

$b=-4$

Let $\log_{5}3125$=c

Index form

$3125=5^c$

$5^5=5^c$

$5=c$

$c=5$

Having gotten the value of all log, let's evaluate together

$\log_{6}216-\log_{¼}256+\log_{5}3125$

$=3-(-4)+5=12$

RELATED POSTS

Example 6

Evaluate

$\log_{7}x=\log_{16}256$

Solution: 

$\log_{7}x=\log_{16}16^2$

$\log_{7}x=2\log_{16}16$

If $\log_{16}16=1$, then

$\log_{7}x=2(1)$

$\log_{7}x=2$

Index form

$x=7^2$

$x=49$

Example 7

Simplify 

$\log_{10}(10+3\sqrt{10})+\log_{10}(10+(\sqrt{90+\sqrt{90}}+\log_{10}(10-(\sqrt{90+\sqrt{90}}))$

Solution:

To simplify this, we multiply the middle log to the right-hand log

$\log_{10}(10+3\sqrt{10})+\log_{10}(10+\sqrt{90+\sqrt{90}})(10-(\sqrt{90+\sqrt{90}}$)

By taking the difference of two squares, this will translates into

$\log_{10}(10+3\sqrt{10})+\log_{10}(100-({\sqrt{90+\sqrt{90}}}){^2}$

Note: Square cancels square root

$\log_{10}(10+3\sqrt{10})+\log_{10}(100-(90+\sqrt{90}$

Now, let's open the bracket 

$\log_{10}(10+3\sqrt{10})+\log_{10}(100-90-\sqrt{90})$

$\log_{10}(10+3\sqrt{10})+\log_{10}(10-\sqrt{9\times10})$

$\log_{10}(10+3\sqrt{10})+\log_{10}(10-3\sqrt{10})$

$\log_{10}(10+3\sqrt{10})(10-3\sqrt{10})$

By taking the difference of two squares, this translates into

$\log_{10}(100-(9(\sqrt{10}^2)$

$\log_{10}(100-(9\times10)$

$\log_{10}(100-90)$

$\log_{10}10=1$

Example 8

Solve $3\log_69+\log_612+\log_664-\log672$

Solution:

$\log_69^3+\log_612+\log_664-log_672$

$\log_6729+\log_612+\log_664-log_672$

$\log_6\frac{729×12×64}{72}$

$\log_67776$

$\log_66^5$

$5\log_66=5(1)=5$

I just introduced logarithm with these 7 examples, In our next posts, you will see some examples that will solidify your knowledge of logarithm, be sure to check it out here.

If you have any questions to ask, you can do so in the comment box. Better still, you can ask our Facebook community

Help us grow our readership by sharing this post

Related Posts

Post a Comment

Subscribe Our Newsletter