Question

the set of exhaustive values of x satisfying 3^(4x)+9^(|x-1|)<=10 is​

Answer 1

Given:

$3^{4x}+9^{|x-1|}<=10$

$[m,log_9(\frac{(\sqrt(n)-1)}{(2)})]$

To find:

m+n=?

Solution:

In the question information is missing so, the correct solution can be defined as follows:

$\to 9^{2x}+9^{(|x-1|)}\leq 10$

The possible values are:

$\to 10= 9^0+9^1 \ or \ 9^0+ 9^0\\ \ 9^1+ 9^0\ \ or \ \ \ 9^0+ 9^0$

$\to 9^{2x}=9^0\ \ \ \ \ \ \ 9^{(|x-1|)}=9^1\\\\\to x=0 \ \ \ _{and} \ \ \ x=2\\\\\\\to 9^{2x}=9^1\ \ \ \ \ \ \ 9^{(|x-1|)}=9^0\\\\\to x=\frac{1}{2}\ \ \ _{and} \ \ \ x=1$

So, the points of x are: $0, \frac{1}{2}, 1, 2$

So, the set values:

$(0,1),(0,2) \ \ \ and \ \ \ (\frac{1}{2},1)$

$m,log_9(\frac{(\sqrt(n)-1)}{(2)})=1\\\\if m=0\\\\\to \frac{(\sqrt(n)-1)}{(2)}=9\\\\\to (\sqrt(n)-1)=18\\\\\to \sqrt(n)=18+1\\\\\to \sqrt(n)=19\\\\\to n=361\\\\m=0 \\\\n=361\\\\\ so,\\\to m+n= 361+0$

               $=361$