Prove that 2(log_10 5-1)=log_10 (1/4)?

Jan 22, 2017

Answer:

$\log \left(\frac{1}{4}\right)$ does not equal 2(log_10 5-1)

Explanation:

Lets start off by evaluating 2(log_10 5-1) By simplifying the expression, we obtained 2log_10 4

According to the properties of logarithmic function,

${\log}_{b} {M}^{p} = p {\log}_{b} M$

Hence 2log_10 4 is equivalent to log_10 4^2 or log_10 16

Another property for logarithmic function is ${\log}_{b} M = {\log}_{b} N$ if and only if M = N

As this equation shows,

$\log \left(\frac{1}{4}\right)$ ≠ $\log 16$ as the values of $M$ and $N$ are not equal to each other.

Jan 22, 2017

Answer:

Yes, $2 \left({\log}_{10} 5 - 1\right) = \log \left(\frac{1}{4}\right)$

Explanation:

Before we seek prove the identity, let us recall a few logarithmic relations.

$\log a - \log b = \log \left(\frac{a}{b}\right)$, $m \log a = \log {a}^{m}$ and ${\log}_{n} n = 1$

As from this we have $1 = {\log}_{10} 10$, we can write

$2 \left({\log}_{10} 5 - 1\right)$

= $2 \left({\log}_{10} 5 - {\log}_{10} 10\right)$

= $2 \left({\log}_{10} \left(\frac{5}{10}\right)\right)$

= $2 \left({\log}_{10} \left(\frac{1}{2}\right)\right)$

= ${\log}_{10} {\left(\frac{1}{2}\right)}^{2}$

= ${\log}_{10} \left(\frac{1}{4}\right)$ or $\log \left(\frac{1}{4}\right)$

As we do not write the base, when using $10$ as base,

we have ${\log}_{10} \left(\frac{1}{4}\right) = \log \left(\frac{1}{4}\right)$