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

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Answer 1 · 2017-01-22T07:07:40

$log (1/4)$ does not equal $2(log_10 5-1)$

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,

$log_bM^p = plog_b M$

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

Another property for logarithmic function is $log_bM = log_b N$ if and only if M = N

As this equation shows,

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

Answer 2 · 2017-01-22T07:37:50

Yes, $2(log_10 5-1)=log (1/4)$

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

$loga-logb=log(a/b)$ , $mloga=log a^m$ and $log_n n=1$

As from this we have $1=log_10 10$ , we can write

$2(log_10 5-1)$

= $2(log_10 5-log_10 10)$

= $2(log_10 (5/10))$

= $2(log_10 (1/2))$

= $log_10 (1/2)^2$

= $log_10 (1/4)$ or $log (1/4)$

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

we have $log_10 (1/4)=log (1/4)$

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