We know that 100 = ${10^2}$, But for what value of K, 100 is equal to ${3^K}$.
So 100 = ${3^K}$ is represented as ${\log _3}100 = K$
Definition: Let a>0, b>0, b$ \ne $1.
$a = {b^x} \Rightarrow {\log _b}a$, logarithm of a to the base b.
Properties:
1. ${\log _b}1 = 0$ ( Logarithm of 1 to any base is 0)
2. ${\log _b}b = 1$ ( Logarithm of b to the same base b is 1)
3. ${\log _b}{b^x}$ = x
4. ${b^{{{\log }_b}x}} = x$
5. ${\log _b}a = {\log _c}a.{\log _b}c$ or ${\log _b}a = \displaystyle\frac{{{{\log }_c}a}}{{{{\log }_c}b}}$
6. ${\log _b}a = \displaystyle\frac{1}{{{{\log }_a}b}}$
7. ${\log _b}xy = {\log _b}x + {\log _b}y$
8. ${\log _b}\displaystyle\frac{x}{y} = {\log _b}x - {\log _b}y$
9. ${\log _b}{a^K} = K.{\log _b}a$
10. ${\log _{{b^K}}}a = \displaystyle\frac{1}{K}.{\log _b}a$
11. ${\log _{{b^K}}}{a^M} = \displaystyle\frac{M}{K}.{\log _b}a$
12. ${a^{{{\log }_c}b}} = {b^{{{\log }_c}a}}$ (a, b, c >0 and c$ \ne $1
Logarithmic inequalities:
1. If ${\rm{0 < b < 1,lo}}{{\rm{g}}_b}x < {\log _b}y \Rightarrow x < y$
2. ${\rm{1 < b,lo}}{{\rm{g}}_b}x < {\log _b}y \Rightarrow x > y$
Solved Examples:
1. a = ${\log _4}5$ and b = ${\log _5}6$, then ${\log _2}3$ =
We have to remove 5 as base and take log to the base 2.
So consider ab = ${\rm{lo}}{{\rm{g}}_4}5.{\log _5}6$ = ${\log _4}6$ = ${\log _{{2^2}}}6$ = $\frac{1}{2}{\log _2}6$
$ \Rightarrow \frac{1}{2}\left( {{{\log }_2}2 \times 3} \right) = \frac{1}{2}\left( {{{\log }_2}2 + {{\log }_2}3} \right) = \displaystyle\frac{1}{2}\left( {1 + {{\log }_2}3} \right)$
So ${{{\log }_2}3}$ = 2ab - 1
2. ${4^{{{\log }_9}3}} + {9^{{{\log }_2}4}} = {10^{{{\log }_x}83}}$ then x =
Sol: Let us first simplify the powers of the two terms on the left hand side
${\log _9}3 = {\log _{{3^2}}}3 = \displaystyle\frac{1}{2}{\log _3}3 = \displaystyle\frac{1}{2}$
Also ${\log _2}4 = {\log _2}{2^2} = 2{\log _2}2 = 1$
Now lefthand side = ${4^{1/2}} + {9^2}$ = 83.
Now ${10^{{{\log }_x}83}} = {83^{{{\log }_x}10}}$
So Equating both sides, 83 = ${83^{{{\log }_x}10}}$ $ \Rightarrow {\log _x}10$ = 1 $ \Rightarrow $ x = 10
3. If ${\log _{12}}27$ = a, then ${\log _6}16$ = ?
So 100 = ${3^K}$ is represented as ${\log _3}100 = K$
Definition: Let a>0, b>0, b$ \ne $1.
$a = {b^x} \Rightarrow {\log _b}a$, logarithm of a to the base b.
Properties:
1. ${\log _b}1 = 0$ ( Logarithm of 1 to any base is 0)
2. ${\log _b}b = 1$ ( Logarithm of b to the same base b is 1)
3. ${\log _b}{b^x}$ = x
4. ${b^{{{\log }_b}x}} = x$
5. ${\log _b}a = {\log _c}a.{\log _b}c$ or ${\log _b}a = \displaystyle\frac{{{{\log }_c}a}}{{{{\log }_c}b}}$
6. ${\log _b}a = \displaystyle\frac{1}{{{{\log }_a}b}}$
7. ${\log _b}xy = {\log _b}x + {\log _b}y$
8. ${\log _b}\displaystyle\frac{x}{y} = {\log _b}x - {\log _b}y$
9. ${\log _b}{a^K} = K.{\log _b}a$
10. ${\log _{{b^K}}}a = \displaystyle\frac{1}{K}.{\log _b}a$
11. ${\log _{{b^K}}}{a^M} = \displaystyle\frac{M}{K}.{\log _b}a$
12. ${a^{{{\log }_c}b}} = {b^{{{\log }_c}a}}$ (a, b, c >0 and c$ \ne $1
Logarithmic inequalities:
1. If ${\rm{0 < b < 1,lo}}{{\rm{g}}_b}x < {\log _b}y \Rightarrow x < y$
2. ${\rm{1 < b,lo}}{{\rm{g}}_b}x < {\log _b}y \Rightarrow x > y$
Solved Examples:
1. a = ${\log _4}5$ and b = ${\log _5}6$, then ${\log _2}3$ =
We have to remove 5 as base and take log to the base 2.
So consider ab = ${\rm{lo}}{{\rm{g}}_4}5.{\log _5}6$ = ${\log _4}6$ = ${\log _{{2^2}}}6$ = $\frac{1}{2}{\log _2}6$
$ \Rightarrow \frac{1}{2}\left( {{{\log }_2}2 \times 3} \right) = \frac{1}{2}\left( {{{\log }_2}2 + {{\log }_2}3} \right) = \displaystyle\frac{1}{2}\left( {1 + {{\log }_2}3} \right)$
So ${{{\log }_2}3}$ = 2ab - 1
2. ${4^{{{\log }_9}3}} + {9^{{{\log }_2}4}} = {10^{{{\log }_x}83}}$ then x =
Sol: Let us first simplify the powers of the two terms on the left hand side
${\log _9}3 = {\log _{{3^2}}}3 = \displaystyle\frac{1}{2}{\log _3}3 = \displaystyle\frac{1}{2}$
Also ${\log _2}4 = {\log _2}{2^2} = 2{\log _2}2 = 1$
Now lefthand side = ${4^{1/2}} + {9^2}$ = 83.
Now ${10^{{{\log }_x}83}} = {83^{{{\log }_x}10}}$
So Equating both sides, 83 = ${83^{{{\log }_x}10}}$ $ \Rightarrow {\log _x}10$ = 1 $ \Rightarrow $ x = 10
3. If ${\log _{12}}27$ = a, then ${\log _6}16$ = ?
