Question

${\log }_{10}(x+1)+lo{g}_{10}(x-1)=lo{g}_{10}11+2lo{g}_{10}3$
The value of x is .

• A. 10
• B. $2{\log }_{10}3$
• C. ${\log }_{10}2$

Answer 1

The correct option is A 10

${\log }_{10}(x+1)+lo{g}_{10}(x-1)=lo{g}_{10}11+2lo{g}_{10}3$
We know ${\log }_{a}m+lo{g}_{a}n=lo{g}_{a}mn$ and
$n{\log }_{a}m=lo{g}_a{m}^n$
Hence
${\log }_{10}(x+1)\times (x-1)=lo{g}_{10}11\times 9$
${\log }_{10}(x^2-1)=lo{g}_{10}99$
Comparing LHS and RHS
$x^2-1=99$
$x^2=100$
$x=10$