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 $\log_{3}27=x$
$27=3^x$
$3^3=3^x$
$3=x$
$x=3$
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
$\log_{6}216-\log_{¼}256+\log_{5}3125$
$=3-(-4)+5=12$
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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.
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