a) Cho $\lg5=a$. Tính $\lg25$ theo $a$.
b) Cho $\lg5=a$. Tính $\lg \frac{1}{64}$ theo $a$.
c) Cho $\lg2=a$. Tính $\lg \frac{125}{4}$ theo $a$.
d) Cho $\log_25=a$. Tính $\log_4500$ theo $a$.
e) Cho $\log_26=a$. Tính $\log_318$ theo $a$.
a) $\lg25=\lg5^2=2\lg5=2\lg(\frac{10}{2})=2(1-\lg2)=2(1-a) $.

b) $\lg \frac{1}{64}=\lg2^{-6}=-6\lg2=-6\lg \frac{10}{5}=-6(1-\lg5)=6(a-1)$.

c) $\lg \frac{125}{4}=\lg(\frac{5^3}{2^2})=\lg5^3-\lg2^2=3\lg5-2\lg2=3(1-\lg2)-2\lg2=3-5a. $

d)$\log_4500=\frac{1}{2}\log_2(2^2.5^3)=\frac{1}{2}(\log_22^2+\log_25^3)=\frac{1}{2}(2+3\log_25)=\frac{1}{2}(2+3a)$.

e) $\log_26=a\Leftrightarrow \log_2(2.3)=a\Leftrightarrow 1+\log_2^3=a$
Suy ra $\log_23=a-1$
$\log_318=\log_3(2.3^2)=\log_32+2=\frac{1}{\log_23}+2=\frac{1}{a-1}+2=\frac{2a-1}{a-1}.$ 

Thẻ

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