logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

The solution of the equation $xlog2+log(1+2^x)-log6=0$

$\begin{array}{1 1}(A)0 \\ (B) 1 \\ (C) 2 \\(D) -1 \end{array}$

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • $loga+logb=log(ab)$
  • $loga-logb=log(\large\frac{a}{b})$
  • $nloga=log(a^n)$
  • $log1=0$
Given: $xlog2+log(1+2^x)-log6=0$
$\Rightarrow\:log(2^x)+log(1+2^x)-log6=0$
$\Rightarrow\:log\bigg(\large\frac{2^x(1+2^x)}{6}\bigg)$$=0$
$\Rightarrow\:\large\frac{2^x(1+2^x)}{6}=1$
$\Rightarrow\:(2^x)^2+2^x-6=0$
Let $2^x=y$
$\Rightarrow\:y^2+y-6=0$
$\Rightarrow\:y=-3\:\:or\:\:2$
$\Rightarrow\:2^x=2\:or\:-3$
But $2^x$ cannot be negative.
$\therefore\:2^x=2\:\:or\:\:x=1$
answered Jul 27, 2013 by rvidyagovindarajan_1
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...