Question

Find the value of $x$, given that $2{\log }_{10}(2^x-1)={\log }_{10}2+{\log }_{10}(2^x+3)$

• A. ${\log }_{2}5$
• B. ${\log }_{10}5$
• C. ${\log }_{4}5$
• D. ${\log }_{2}7$

Answer 1

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

Given,

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

let $2^x=u$

$2{\log }_{10}(u-1)={\log }_{10}\left(2\right)+{\log }_{10}(u+3)$

$2{\log }_{10}(u-1)={\log }_{10}\left(2(u+3)\right)$

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

${(u-1)}^2=2(u+3)$

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

$u^2-4u-5=0$

$(u-5)(u+1)=0$

$u=-1,5$

$\Rightarrow 2^x=-1,5$

since, $\log (-1)=undefined$,
So, $\log 2^x=\log 5$

$x\log 2=\log 5$

$x=\frac{\log 5}{\log 2}$

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