Question

If log 8 = 0.9030, find the value of
(b) log 2000
(a) log 64

Answer 1

We have the identities,

• $\log(ab)=\log a+\log b\quad\quda\dots(i)$

• $\log(a^b)=b\log a\quad\quad\dots(ii)$

Given,

$\longrightarrow \log8=0.9030$

$\longrightarrow \log(2^3)=0.9030$

By (ii),

$\longrightarrow 3\log2=0.9030$

$\longrightarrow \log2=\dfrac{0.9030}{3}$

$\longrightarrow \log2=0.3010$

Here,

$\longrightarrow \log(2000)=\log(2\times1000)$

$\longrightarrow \log(2000)=\log(2\times10^3)$

By (i),

$\longrightarrow \log(2000)=\log(2)+\log(10^3)$

$\longrightarrow \log(2000)=0.3010+\log(10^3)$

By (ii),

$\longrightarrow \log(2000)=0.3010+3\log10$

Since $\log10=1,$

$\longrightarrow \log(2000)=0.3010+3\times1$

$\longrightarrow \log(2000)=0.3010+3$

$\longrightarrow\underline{\underline{\log(2000)=3.3010}}$

And,

$\longrightarrow\log(64)=\log(8^2)$

$\longrightarrow\log(64)=2\log(8)$

$\longrightarrow\log(64)=2\times0.9030$

$\longrightarrow\underline{\underline{\log(64)=1.8060}}$