Question

If $2{\log }_{10}(2^x-1)={\log }_{10}2+{\log }_{10}(2^x+3)$, then the value of $x$ will be

• A. ${\log }_{2}5$
• B. ${\log }_{5}2$
• C. ${\log }_{2}2$
• D. ${\log }_{5}4$

Answer 1

The correct option is A ${\log }_{2}5$

Given:

$2{\log }_{10}{(2}^x-1)={\log }_{10}2+{\log }_{10}{(2}^x+3)$

Let assume $2^x=a,a>1$

$2{\log }_{10}(a-1)={\log }_{10}2+{\log }_{10}(a+3)$

Using logarithmic properties, we get

$\log ({a-1)}^2=\log 2(a+3)$

Removing log from both sides,

$a^2+1-2a=2a+6$

$\Rightarrow a^2-4a-5=0$

$\Rightarrow (a-5)(a+1)=0a=5$

$\therefore 2^x=5$

$a$ cannot be $-1$ because $a=2^x$ cannot be negative.

Taking log on both sides

$\therefore {\log }_{2}5=x$

Hence, option A.