Prove that $2 ({log}_{10} 5 - 1) = {log}_{10} (\frac{1}{4})$ $2(log_10 5-1)=log_10 (1/4)$ ?

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Answer 1 · 2017-01-22T07:07:40

$log (\frac{1}{4})$ $log (1/4)$ does not equal $2 ({log}_{10} 5 - 1)$ $2(log_10 5-1)$

Lets start off by evaluating $2 ({log}_{10} 5 - 1)$ $2(log_10 5-1)$ By simplifying the expression, we obtained $2 {log}_{10} 4$ $2log_10 4$

According to the properties of logarithmic function,

${log}_{b} M^{p} = p {log}_{b} M$ $log_bM^p = plog_b M$

Hence $2 {log}_{10} 4$ $2log_10 4$ is equivalent to ${log}_{10} 4^{2} or {log}_{10} 16$ $log_10 4^2 or log_10 16$

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

As this equation shows,

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

Answer 2 · 2017-01-22T07:37:50

Yes, $2 ({log}_{10} 5 - 1) = log (\frac{1}{4})$ $2(log_10 5-1)=log (1/4)$

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

$log a - log b = log (\frac{a}{b})$ $loga-logb=log(a/b)$ , $m log a = {log a}^{m}$ $mloga=log a^m$ and ${log}_{n} n = 1$ $log_n n=1$

As from this we have $1 = {log}_{10} 10$ $1=log_10 10$ , we can write

$2 ({log}_{10} 5 - 1)$ $2(log_10 5-1)$

= $2 ({log}_{10} 5 - {log}_{10} 10)$ $2(log_10 5-log_10 10)$

= $2 ({log}_{10} (\frac{5}{10}))$ $2(log_10 (5/10))$

= $2 ({log}_{10} (\frac{1}{2}))$ $2(log_10 (1/2))$

= ${log}_{10} {(\frac{1}{2})}^{2}$ $log_10 (1/2)^2$

= ${log}_{10} (\frac{1}{4})$ $log_10 (1/4)$ or $log (\frac{1}{4})$ $log (1/4)$

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

we have ${log}_{10} (\frac{1}{4}) = log (\frac{1}{4})$ $log_10 (1/4)=log (1/4)$

Prove that 2(log105−1)=log10(14)2(log_10 5-1)=log_10 (1/4)?