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Basic properties of logarithms

The logarithm of a given number b is the exponent to which another fixed number, the base a, must be raised, to produce that number b $(a>0, \ a\neq 1, \ b>0)$.

$$\displaystyle \log_ab=c, \ a^c=b$$

Logarithmic identities

Logarithm power rules are very useful whenever expressions involving logarithmic functions need to be simplified:

$$\displaystyle \alpha \log _ax=\log _a x^{\alpha}$$ $$\frac{1}{\alpha}\log _ax=\log _{a^{\alpha}}x$$ $$\displaystyle a^{\log_ax}=x$$

The base of logarithm can be changed in other. It's very helpful, when we need to calculate logarithm with calculator which has only standard 10 and e bases. Change of base formula:

$$\displaystyle \log_ax=\frac{\log_bx}{\log_ba}$$

Sum and difference of two logarithms with the same base properties (or product to sum and quotient to difference) formulas:

$$\log_ax+\log_ay=\log_axy$$ $$\log_ax-\log_ay=\log_a\frac{x}{y}$$

Solved examples

Solve this expression $\log_2{8}+3\log_3{9}$

$$\log_2{8}+3\log_3{9}=3+3\cdot 2=9$$

Simplify this expression $2\log_2{3}+\log_2{9}-2\log_2{27}$

$$2\log_2{3}+\log_2{9}-2\log_2{27}=2\log_2{3}+\log_2{3^2}-2\log_2{3^3}=2\log_2{3}+2\log_2{3}-2\cdot 3 \log_2{3}=-2\log_2{3}$$

Solve this expression $27^{\log_3{5}}+25^{\log_5{2}}$


Convert $\log_2{5}$ to an expression with logarithms having a base of 10.


Solve this expression $\log_2{3}+\log_2{6}-\log_2{9}$

$$\log_2{3}+\log_2{6}-\log_2{9}=\log_2{\frac{3\cdot 6}{9}}=\log_2{2}=1$$

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